Trigonometric equations practice · Medium

Option A ($\pi / 6$) is correct because substituting $x = \pi/6$ into the equation $81^{\sin^2 x} + 81^{\cos^2 x} = 30$ satisfies the equation.

Option B ($2 n \pi+\frac{{4 \pi}}{{3}}$) is the correct general solution because it satisfies both $\sin \theta=-\frac{{\sqrt{{3}}}}{{2}}$ and $\tan \theta=\sqrt{{3}}$.

Option A ($\frac{{2 b}}{{c+a}}, \frac{{c-a}}{{c+a}}$) is correct because it gives the correct expressions for $\tan \alpha + \tan \beta$ and $\tan \alpha \tan \beta$ based on the given equation.

Option D ($-1+4 \cos \frac{{\mathrm{{A}}}}{{2}} \cos \frac{{\mathrm{{B}}}}{{2}} \sin \frac{{\mathrm{{C}}}}{{2}}$) is correct because it represents the simplified form of $\cos A + \cos B - \cos C$ when $A + B + C = \pi$, using trigonometric identities.

Option B ($\left[-\frac{3}{2},-1\right]$) is correct because it represents the range of the given trigonometric expression after simplification and using the range of $\cos^2 2x$.

Option A ([-3,-2]) is correct because by solving the quadratic equation for $\cos^2 x$ and using the fact that $0 \leq \cos^2 x \leq 1$, we find that $-3 \leq a \leq -2$ for the equation to possess a solution.

Option A ($\frac{\pi}{4}, \frac{\pi}{3}, \frac{5 \pi}{12}$) is correct because by using the given condition and the properties of an arithmetic progression, we can find the values of A, B, and C.

Option A is correct because it represents the quadratic equation whose roots are $\tan x$ and $\tan y$, derived from the given conditions $\cot x \cot y = k$ and $x + y = \pi/3$.

Option C ($(2 n+1) \frac{\pi}{2}, n \in I$) is the correct solution because it satisfies both $\sin^6 x = 1$ and $\cos^4 3x = 0$.

Option C (6) is correct because solving the equation $\tan x + \cot x = 2 \csc x$ leads to $\cos x = \frac{1}{2}$, which has 6 solutions in the interval $[-2\pi, 2\pi]$.

Option C ($2 n \pi+\frac{{7 \pi}}{{6}}$) is correct because it represents the general solution that satisfies both the sine and tangent equations.

Option B ($2 {n} \pi+\frac{{2 \pi}}{{3}}$) is correct because it represents the general solution to the trigonometric equation after simplification and solving for x.

Option A ($a$) is correct because by substituting $\tan x = \frac{{b}}{{a}}$ into the expression $a \cos 2x + b \sin 2x$ and using trigonometric identities, the expression simplifies to $a$.

Option B ($2 n \pi \pm \frac{{5 \pi}}{{6}}$) is correct because it represents the general solution derived from solving the given trigonometric equation.

The correct answer is $|\sec x| < 1$ because the range of $\sec x$ is $(-\infty, -1] \cup [1, \infty)$, so its absolute value cannot be less than 1.

Option D (A straight line passing through the point $\left(\frac{\pi}{2},-\sin ^2 1\right)$ and parallel to the $X$-axis) is correct because the given expression simplifies to a constant value, $-\sin^2 1$, which represents a horizontal line.

Option A ($x = \frac{{2\pi }}{{3}}\,\,\;\mathrm{or}\,\,\frac{\pi }{3}$) is correct because it satisfies the given infinite geometric series equation after solving for sin(x).

The correct answer is 9 because the sum of the cosines of the solutions plus the number of solutions to the trigonometric equation equals 9.

Option B (1) is correct because it simplifies to $\sin^2(3\pi) + \cos^2(4\pi) + \tan^2(5\pi) = 0 + 1 + 0 = 1$.

Option C ($2 {n} \pi$) is correct because it represents the general solution to the trigonometric equation after simplification and considering the range of cosine function.

Option B ($\sqrt{1+\sin ^2 \theta}$) is correct because by completing the square and using trigonometric identities, we can find the maximum value of $f(x)$.

Option C ($\frac{{\pi}}{{6}}$) is correct because it is the only valid solution within the given interval that satisfies the equation after checking for extraneous solutions.

Option C ($\frac{3}{2}$) is correct because by applying the sine addition formula and componendo and dividendo, the ratio $\frac{\tan (A+B)}{\tan A}$ simplifies to $\frac{3}{2}$.

Option B ({π/3, 5π/3}) is correct because the equation (2cos(x)-1)(3+2cos(x))=0 implies cos(x) = 1/2 or cos(x) = -3/2. Since cos(x) cannot be -3/2, cos(x) = 1/2, which gives x = π/3 and x = 5π/3 in the given interval.

Option A ($\{x \mid x=2 n \pi \pm \frac{\pi}{3} ; n \in I\}$) is correct because it represents the values of alpha that satisfy the given maximum condition after simplifying the trigonometric expression.

The correct answer is None of these because the expression simplifies to $\frac{1}{2^{n-1}}$, which is not among the options.

Option A ($\cos A \cos B=\frac{1}{5}$) is correct because by using the given trigonometric identities and solving for $\cos A \cos B$, we arrive at the value of $\frac{1}{5}$.

Option B, 2, is correct because there are two solutions to the given equation within the specified interval.

Option A ($\theta=n \pi+(-1)^n\left(\frac{{-3 \pi}}{{10}}\right)$) is correct because by factoring the equation and solving for sin theta, we can find the general solution for theta.

Option A ($b-a=1+a b$) is correct because by substituting the given expressions and simplifying, we can derive the relation $b-a=1+ab$.