Trigonometric Ratios And Identities – JEE Main MathematicsPractice Questions & PYQs

The equation $\cos 2x + a \sin x = 2a - 7$ is solved by applying the identity $\cos 2x = 1 - 2 \sin^2 x$. Substituting this leads to $1 - 2 \sin^2 x + a \sin x = 2a - 7$, which rearranges to $a (\sin x - 2) = 2 (\sin^2 x - 4)$. Since $\sin^2 x - 4 = (\sin x - 2)(\sin x + 2)$, we express this as $a (\sin x - 2) = 2 (\sin x - 2) (\sin x + 2)$. Because $\sin x$ cannot equal 2, we divide both sides by $(\sin x - 2)$ to find $a = 2 (\sin x + 2)$. Given that the range of $\sin x$ is constrained between -1 and 1, the lower bound for $a$ is $2(-1+2) = 2$ and the upper bound is $2(1+2) = 6$, confirming that $p=2$ and $q=6$.

The expression $r = \tan 9^{\circ} - \tan 27^{\circ} - \frac{1}{\cot 63^{\circ}} + \tan 81^{\circ}$ is simplified by using the reciprocal identity $\frac{1}{\cot 63^{\circ}} = \tan 63^{\circ}$ and the co-function identity $\tan 81^{\circ} = \cot 9^{\circ}$. By grouping these, $r$ becomes $(\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \tan 63^{\circ})$. Since $\tan 63^{\circ} = \cot 27^{\circ}$, this is $(\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$. Applying the identity $\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$, the terms transform into $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$. Substituting $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$ results in $8 \left( \frac{1}{\sqrt{5}-1} - \frac{1}{\sqrt{5}+1} \right)$, which simplifies to $8 \left( \frac{\sqrt{5}+1 - (\sqrt{5}-1)}{4} \right) = 4$. Consequently, the product $pqr = 2 \times 6 \times 4 = 48$.

Substituting the exact trigonometric values $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$ and $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ into the given expression allows for direct algebraic simplification. This substitution results in the following calculation:
$\frac{3 \left( \frac{\sqrt{5}+1}{4} \right) + 5 \left( \frac{\sqrt{5}-1}{4} \right)}{5 \left( \frac{\sqrt{5}+1}{4} \right) - 3 \left( \frac{\sqrt{5}-1}{4} \right)} = \frac{8\sqrt{5}-2}{2\sqrt{5}+8}$
Dividing both the numerator and the denominator by two yields $\frac{4\sqrt{5}-1}{\sqrt{5}+4}$. To rationalize the denominator, multiply both parts by the conjugate $\sqrt{5}-4$ to obtain:
$\frac{(4\sqrt{5}-1)(\sqrt{5}-4)}{(\sqrt{5}+4)(\sqrt{5}-4)} = \frac{20 - 16\sqrt{5} - \sqrt{5} + 4}{5-16} = \frac{24-17\sqrt{5}}{-11}$
This expression simplifies to $\frac{17\sqrt{5}-24}{11}$, identifying the values $a=17$, $b=24$, and $c=11$. Adding these natural numbers together results in a final sum of $17 + 24 + 11 = 52$.

Finding the value of $A+B$ begins with the tangent sum formula $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$. Substituting the given expressions, the numerator becomes $\frac{1}{\sqrt{x(x^2+x+1)}} + \frac{\sqrt{x}}{\sqrt{x^2+x+1}}$, which simplifies to $\frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}}$. Simultaneously, the denominator evaluates to $1 - \left(\frac{1}{\sqrt{x(x^2+x+1)}}\right)\left(\frac{\sqrt{x}}{\sqrt{x^2+x+1}}\right)$, or $1 - \frac{1}{x^2+x+1}$, which equals $\frac{x^2+x}{x^2+x+1}$.

Dividing the numerator by the denominator simplifies to $\frac{1+x}{\sqrt{x}\sqrt{x^2+x+1}} \cdot \frac{x^2+x+1}{x(1+x)}$, which further reduces to $\frac{\sqrt{x^2+x+1}}{x\sqrt{x}}$. This expression is identical to $\tan C$, as $\tan C = \sqrt{x^{-3}+x^{-2}+x^{-1}} = \sqrt{\frac{1+x+x^2}{x^3}} = \frac{\sqrt{x^2+x+1}}{x\sqrt{x}}$. Given the constraints on the angles, the equality $\tan(A+B) = \tan C$ leads directly to the conclusion that $A+B=C$.

Transforming a trigonometric equation into an algebraic form is best achieved by applying the half-angle substitution $t = \tan\frac{\theta}{2}$. Utilizing the identities $\cos\theta = \frac{1-t^2}{1+t^2}$ and $\sin\theta = \frac{2t}{1+t^2}$ allows the equation $4\cos\theta - 3\sin\theta = 1$ to be rewritten by substituting these terms, which leads to the quadratic relation:
$5t^2 + 6t - 3 = 0$
Solving for $t$ via the quadratic formula yields two possible values, but the restriction $\theta \in \left[0, \frac{\pi}{4}\right]$ requires selecting the positive root $t = \frac{2\sqrt{6}-3}{5}$. Substituting this value back into the expression $\cos\theta = \frac{1-t^2}{1+t^2}$ requires calculating the ratio of these terms. After performing the algebraic simplification, the value of $\cos\theta$ is $\frac{4}{3\sqrt{6}-2}$.

The expression $\cos^2 2x - 2 \sin^4 x - 2 \cos^2 x = \lambda$ simplifies to a quadratic form by expressing all terms in $\cos^2 x$. Using the identities $\cos 2x = 2\cos^2 x - 1$ and $\sin^2 x = 1 - \cos^2 x$, the equation becomes $(2 \cos^2 x - 1)^2 - 2(1 - \cos^2 x)^2 - 2 \cos^2 x$, which reduces to $2\cos^4 x - 2\cos^2 x - 1$. By setting $t = \cos^2 x$ with the domain $t \in [0, 1]$, we analyze the function $f(t) = 2t^2 - 2t - 1$, which completes to $2(t - \frac{1}{2})^2 - \frac{3}{2}$. Since the vertex at $t = \frac{1}{2}$ is within the domain, the function achieves a minimum value of $-\frac{3}{2}$ and reaches a maximum of $-1$ at the interval endpoints $t=0$ and $t=1$, defining the range of $\lambda$ as $\left[ -\frac{3}{2}, -1 \right]$.

Each term in the trigonometric expression simplifies based on standard values and periodicity, allowing us to combine them into a single constant. Recalling that $\tan 15^{\circ} = 2-\sqrt{3}$, we note that $\frac{1}{\tan 75^{\circ}}$ is equivalent to $\cot 75^{\circ}$, which equals $\tan 15^{\circ}$, or $2-\sqrt{3}$. Similarly, $\frac{1}{\tan 105^{\circ}}$ equates to $\cot 105^{\circ}$, which simplifies to $-\cot 75^{\circ}$, or $\sqrt{3}-2$, while $\tan 195^{\circ}$ reduces to $\tan 15^{\circ}$ because the tangent function has a period of $180^{\circ}$, giving $2-\sqrt{3}$. Summing these values yields $(2-\sqrt{3}) + (2-\sqrt{3}) + (\sqrt{3}-2) + (2-\sqrt{3}) = 4-2\sqrt{3}$. Setting $2a = 4-2\sqrt{3}$ identifies $a = 2-\sqrt{3}$, and computing the reciprocal $\frac{1}{a} = \frac{1}{2-\sqrt{3}} = 2+\sqrt{3}$ makes the final sum $a+\frac{1}{a} = (2-\sqrt{3}) + (2+\sqrt{3}) = 4$.

The expression is simplified by grouping terms with complementary angles, observing that $\tan 81^{\circ} = \cot 9^{\circ}$ and $\tan 63^{\circ} = \cot 27^{\circ}$, which transforms the problem into $(\tan 9^{\circ} + \cot 9^{\circ}) - (\tan 27^{\circ} + \cot 27^{\circ})$. Utilizing the trigonometric identity $\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}$ allows this to be rewritten as $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}}$. Substituting the known values $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\sin 54^{\circ} = \frac{\sqrt{5}+1}{4}$ results in $2 \left( \frac{4}{\sqrt{5}-1} - \frac{4}{\sqrt{5}+1} \right)$, which simplifies to $8 \left( \frac{(\sqrt{5}+1) - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} \right)$. Completing the calculation yields $8 \left( \frac{2}{5-1} \right)$, which simplifies to $8 \left( \frac{2}{4} \right)$, resulting in 4.

The trigonometric identity $4 \cos^2 \theta - 1 = \frac{\sin 3\theta}{\sin \theta}$ transforms the provided expression into a telescoping product. This relation arises because $4 \cos^2 \theta - 1 = 4(1 - \sin^2 \theta) - 1 = 3 - 4 \sin^2 \theta$, which is equivalent to $\frac{\sin 3\theta}{\sin \theta}$ since $\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta$.

Substituting the angles $9^{\circ}, 27^{\circ}, 81^{\circ},$ and $243^{\circ}$ into this identity results in:
$36 \times \frac{\sin 27^{\circ}}{\sin 9^{\circ}} \times \frac{\sin 81^{\circ}}{\sin 27^{\circ}} \times \frac{\sin 243^{\circ}}{\sin 81^{\circ}} \times \frac{\sin 729^{\circ}}{\sin 243^{\circ}}$
Most terms cancel out as the product unfolds, leaving $36 \times \frac{\sin 729^{\circ}}{\sin 9^{\circ}}$. Because $729^{\circ}$ corresponds to $2 \times 360^{\circ} + 9^{\circ}$, the expression simplifies to $36 \times \frac{\sin 9^{\circ}}{\sin 9^{\circ}}$, which equals 36.

To simplify the given function $f(\theta)$, we first utilize trigonometric reduction identities, noting that $\sin^4(\frac{3\pi}{2}-\theta)$ is equivalent to $\cos^4\theta$ and $\sin^4(3\pi+\theta)$ is equal to $\sin^4\theta$. Using the identity $\cos^4\theta + \sin^4\theta = 1 - \frac{1}{2}\sin^2 2\theta$, we transform the initial expression $f(\theta)=3(\cos^4\theta+\sin^4\theta)-2\cos^2 2\theta$. Substituting this into the function yields $f(\theta) = 3(1 - \frac{1}{2}\sin^2 2\theta) - 2\cos^2 2\theta$. Since $\sin^2 2\theta = 1 - \cos^2 2\theta$, the function simplifies further to $f(\theta) = \frac{3}{2} - \frac{1}{2}\cos^2 2\theta$. Applying the double-angle formula $\cos^2 2\theta = \frac{1+\cos 4\theta}{2}$ allows us to express $f(\theta)$ as $\frac{5}{4} - \frac{\cos 4\theta}{4}$.

Differentiating with respect to $\theta$ gives $f'(\theta) = \sin 4\theta$. We determine the set $S$ by solving $\sin 4\theta = -\frac{\sqrt{3}}{2}$ for $\theta \in [0, \pi]$. Summing the solutions in the interval results in $4\beta = \frac{3\pi}{2}$. Consequently, evaluating the simplified expression at $\beta$ results in $f(\beta) = \frac{5}{4} - \frac{1}{4}\cos(\frac{3\pi}{2})$, which simplifies to $\frac{5}{4}$.

The product of a cosine series where each consecutive angle is double the previous one follows the general identity $\prod_{k=0}^{n-1} \cos(2^k A) = \frac{\sin(2^n A)}{2^n \sin A}$. For this specific problem, the first angle $A = \frac{\pi}{33}$ and there are $n = 5$ terms, corresponding to the angles $\frac{\pi}{33}, \frac{2\pi}{33}, \frac{4\pi}{33}, \frac{8\pi}{33},$ and $\frac{16\pi}{33}$.

Applying these values to the formula transforms the expression into $96 \times \frac{\sin(32 \cdot \frac{\pi}{33})}{32 \sin(\frac{\pi}{33})}$. Dividing $96$ by $32$ yields $3$, and the remaining sine ratio simplifies because $\sin(\frac{32\pi}{33})$ is equivalent to $\sin(\pi - \frac{\pi}{33})$, which equals $\sin(\frac{\pi}{33})$. Since the sine terms in the numerator and denominator cancel each other out, the expression simplifies to 3.

The triple angle identity $\sin \theta \sin(60^{\circ}-\theta) \sin(60^{\circ}+\theta) = \frac{1}{4} \sin(3\theta)$ is the primary tool for simplifying products of sines. We can group the original expression as $(\sin 10^{\circ} \sin 50^{\circ} \sin 70^{\circ}) \times (\sin 10^{\circ} \sin 20^{\circ} \sin 40^{\circ})$. The first group matches the triple angle identity with $\theta = 10^{\circ}$, simplifying directly to $\frac{1}{4} \sin(30^{\circ}) = \frac{1}{8}$.

For the second group, applying the product-to-sum identity $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ to $\sin 20^{\circ} \sin 40^{\circ}$ results in $\frac{1}{2}(\cos 20^{\circ} - \cos 60^{\circ})$, or $\frac{1}{2}(\cos 20^{\circ} - \frac{1}{2})$. Multiplying this by the remaining $\sin 10^{\circ}$ yields $\frac{1}{2} \sin 10^{\circ} \cos 20^{\circ} - \frac{1}{4} \sin 10^{\circ}$. Combining the two groups gives $\frac{1}{8} \times (\frac{1}{2} \sin 10^{\circ} \cos 20^{\circ} - \frac{1}{4} \sin 10^{\circ})$, which simplifies to $\frac{1}{16} \sin 10^{\circ} \cos 20^{\circ} - \frac{1}{32} \sin 10^{\circ}$. By substituting $2 \sin 10^{\circ} \cos 20^{\circ} = \sin 30^{\circ} - \sin 10^{\circ}$, the expression becomes $\frac{1}{32}(\sin 30^{\circ} - \sin 10^{\circ}) - \frac{1}{32} \sin 10^{\circ}$. With $\sin 30^{\circ} = \frac{1}{2}$, this reduces to $\frac{1}{64} - \frac{1}{16} \sin 10^{\circ}$. Comparing this to the given form $\alpha - \frac{1}{16} \sin 10^{\circ}$, we find $\alpha = \frac{1}{64}$, making $16 + \alpha^{-1} = 16 + 64 = 80$.

Trigonometric reduction relies on the effective application of sum-to-product identities. Rewriting $2 \sin 12^{\circ}$ as $\sin 12^{\circ} + \sin 12^{\circ}$ allows the expression to be grouped as $\sin 12^{\circ} + (\sin 12^{\circ} - \sin 72^{\circ})$. Applying the identity $\sin A - \sin B = 2 \sin \left( \frac{A-B}{2} \right) \cos \left( \frac{A+B}{2} \right)$ to the grouped term produces $2 \sin(-30^{\circ}) \cos 42^{\circ}$, which simplifies to $\sin 12^{\circ} - \cos 42^{\circ}$ since $\sin(-30^{\circ}) = -1/2$. By converting the cosine term into its co-function $\sin 48^{\circ}$, the expression becomes $\sin 12^{\circ} - \sin 48^{\circ}$. Repeating the difference-to-product identity yields $2 \sin(-18^{\circ}) \cos 30^{\circ}$. Substituting the standard values $-2 \cdot \left( \frac{\sqrt{5}-1}{4} \right) \cdot \frac{\sqrt{3}}{2}$ simplifies the final result to $\frac{\sqrt{3}(1-\sqrt{5})}{4}$.

Evaluating the product of these sine terms relies on the complementary angle identity $\sin(\theta) = \cos(\frac{\pi}{2} - \theta)$ to simplify the arguments. By applying this identity, each factor $\sin(\frac{k\pi}{22})$ converts to $\cos(\frac{(11-k)\pi}{22})$, which rewrites the entire expression as:

$2 \cos\left(\frac{\pi}{11}\right) \cos\left(\frac{2\pi}{11}\right) \cos\left(\frac{3\pi}{11}\right) \cos\left(\frac{4\pi}{11}\right) \cos\left(\frac{5\pi}{11}\right)$

Utilizing the general product identity $\prod_{k=1}^n \cos(\frac{k\pi}{2n+1}) = \frac{1}{2^n}$ with $n=5$, the product of these five cosine terms equals $\frac{1}{2^5}$, which simplifies to $\frac{1}{32}$. Multiplying this result by the initial factor of $2$ yields $2 \times \frac{1}{32} = \frac{1}{16}$.

Identifying $\tan\alpha$ and $\tan\beta$ is the first step in solving for $\tan(\alpha+\beta)$. Since $\cot\alpha=1$, it follows that $\tan\alpha=1$, and because $\sec\beta=-\frac{5}{3}$ in the second quadrant, the trigonometric ratios yield $\tan\beta=-\frac{4}{3}$. Applying the identity $\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$ results in $\frac{1-\frac{4}{3}}{1-(1)(-\frac{4}{3})}$, which simplifies to $-\frac{1}{7}$.

The range of $\alpha+\beta$ is determined by adding the boundaries of the given intervals, $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$. This addition yields $\frac{3\pi}{2} < \alpha+\beta < \frac{5\pi}{2}$, with the angle specifically landing in the fourth quadrant.

The value of $\sin 36^{\circ}$ is known to be $\frac{\sqrt{10-2\sqrt{5}}}{4}$. Squaring both sides of the equation $x = \frac{\sqrt{10-2\sqrt{5}}}{4}$ provides $x^2 = \frac{10-2\sqrt{5}}{16}$, which rearranges into $16x^2 = 10 - 2\sqrt{5}$. Isolating the remaining radical term leads to the expression $16x^2 - 10 = -2\sqrt{5}$. When both sides are squared again to eliminate the square root, the result is $(16x^2 - 10)^2 = (-2\sqrt{5})^2$. Expanding this yields $256x^4 - 320x^2 + 100 = 20$. Subtracting 20 from both sides produces $256x^4 - 320x^2 + 80 = 0$, and dividing the entire equation by 16 simplifies the expression to $16x^4 - 20x^2 + 5 = 0$.

The sum of cosines with angles in an arithmetic progression follows the formula $\frac{\sin(n\delta/2)}{\sin(\delta/2)} \cdot \cos\left(\frac{\theta_1 + \theta_n}{2}\right)$, where $n$ represents the number of terms, $\delta$ is the common difference, and $\theta_1$ and $\theta_n$ are the first and last angles. In the series $\cos\left(\frac{2\pi}{7}\right)+\cos\left(\frac{4\pi}{7}\right)+\cos\left(\frac{6\pi}{7}\right)$, there are three terms with a common difference $\delta = \frac{2\pi}{7}$, resulting in the expression $\frac{\sin\left(\frac{3\pi}{7}\right)}{\sin\left(\frac{\pi}{7}\right)} \cdot \cos\left(\frac{4\pi}{7}\right)$. Multiplying the numerator and denominator by two allows the use of the trigonometric identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$, which expands the numerator to $\sin\left(\frac{3\pi}{7} + \frac{4\pi}{7}\right) + \sin\left(\frac{3\pi}{7} - \frac{4\pi}{7}\right)$. Since $\sin\left(\frac{7\pi}{7}\right) = \sin(\pi) = 0$ and $\sin\left(-\frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)$, the calculation yields $\frac{-\sin\left(\frac{\pi}{7}\right)}{2\sin\left(\frac{\pi}{7}\right)}$, which simplifies to $-\frac{1}{2}$.

The core concept behind evaluating this product lies in the trigonometric identity:
$4 \sin \theta \sin(60^{\circ} - \theta) \sin(60^{\circ} + \theta) = \sin 3\theta$
By setting $\theta$ to $20^{\circ}$, the terms in the expression align perfectly with this identity because $60^{\circ} - 20^{\circ}$ is $40^{\circ}$ and $60^{\circ} + 20^{\circ}$ is $80^{\circ}$. We can rewrite the original expression as 4 multiplied by the grouping $4 \sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}$. Applying the identity allows us to substitute that grouping with $\sin(3 \times 20^{\circ})$, which simplifies to $\sin 60^{\circ}$. Multiplying this result by 4 gives $4 \sin 60^{\circ}$, and since $\sin 60^{\circ}$ equals $\frac{\sqrt{3}}{2}$, the final value is $2\sqrt{3}$.

The properties of logarithms allow us to combine terms efficiently, while the fundamental Pythagorean identity of trigonometry provides the necessary link between the sum and product of sine and cosine functions. Starting with the first given equation, we utilize the product rule for logarithms to obtain:

$\log_{10}(\sin x \cos x) = -1$

This implies that the product $\sin x \cos x$ equals 0.1.

For the second equation, we can multiply both sides by 2 to eliminate the fraction, resulting in $2 \log_{10}(\sin x + \cos x) = \log_{10} n - \log_{10} 10$. Applying the power rule of logarithms allows us to rewrite this as $\log_{10}(\sin x + \cos x)^2 = \log_{10}(\frac{n}{10})$. Removing the logarithms from both sides gives the identity:

$(\sin x + \cos x)^2 = \frac{n}{10}$

Expanding the left side yields $\sin^2 x + \cos^2 x + 2 \sin x \cos x = \frac{n}{10}$. Since the Pythagorean identity tells us that $\sin^2 x + \cos^2 x = 1$, we substitute this and our known value for $\sin x \cos x$ to find $1 + 2(0.1) = \frac{n}{10}$. This simplifies to $1.2 = \frac{n}{10}$, confirming that $n = 12$.

The equation $15 \sin^{4}\alpha + 10 \cos^{4}\alpha = 6$ can be unified using the fundamental identity $\sin^{2}\alpha + \cos^{2}\alpha = 1$. By multiplying the right side by $(\sin^{2}\alpha + \cos^{2}\alpha)^{2}$, the equation becomes homogeneous in terms of trigonometric powers. Expanding the identity on the right leads to $15 \sin^{4}\alpha + 10 \cos^{4}\alpha = 6(\sin^{4}\alpha + 2 \sin^{2}\alpha \cos^{2}\alpha + \cos^{4}\alpha)$, which allows for algebraic simplification.

Rearranging the terms results in $9 \sin^{4}\alpha - 12 \sin^{2}\alpha \cos^{2}\alpha + 4 \cos^{4}\alpha = 0$. This expression represents a perfect square, specifically $(3 \sin^{2}\alpha - 2 \cos^{2}\alpha)^{2} = 0$, meaning that $3 \sin^{2}\alpha$ must equal $2 \cos^{2}\alpha$. This relationship establishes that $\tan^{2}\alpha = \frac{2}{3}$, and reciprocally, $\cot^{2}\alpha = \frac{3}{2}$.

With these values, the expression $27 \sec^{6}\alpha + 8 \csc^{6}\alpha$ can be evaluated by converting secant and cosecant into tangent and cotangent identities. Substituting $\sec^{2}\alpha = 1 + \tan^{2}\alpha$ and $\csc^{2}\alpha = 1 + \cot^{2}\alpha$ transforms the expression into $27(1 + \tan^{2}\alpha)^{3} + 8(1 + \cot^{2}\alpha)^{3}$. Inserting the previously derived ratios provides $27(1 + \frac{2}{3})^{3} + 8(1 + \frac{3}{2})^{3}$, which simplifies to $27(\frac{5}{3})^{3} + 8(\frac{5}{2})^{3}$. Evaluating these terms leads to $125 + 125$, confirming the final value is 250.

The expression $a \sin x + b \cos x$ always produces values in the range $[-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]$. By applying this property to the equation $3 \sin x + 4 \cos x = k + 1$, we find that the left side oscillates between $-\sqrt{3^2 + 4^2}$ and $\sqrt{3^2 + 4^2}$, which evaluates to the closed interval $[-5, 5]$. A solution for $x$ exists only when the value of the right side, $k + 1$, also falls within this interval. Consequently, we require $-5 \leq k + 1 \leq 5$, which simplifies to $-6 \leq k \leq 4$ upon subtracting 1 from each part of the inequality. The valid integers for $k$ are -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, and 4, resulting in a total of 11 possible values.