Trigonometric equations practice · Mixed difficulty

Option A ($n \pi+(-1)^{{n+1}} \frac{{\pi}}{{6}}$) is correct because it represents the general solution derived from solving the given trigonometric equation.

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 (18) is correct because applying AM >= GM inequality, the minimum value of the expression is 18.

The correct answer is C ($\pi / 3$) because substituting $x = \sin \theta$ and simplifying the expression leads to this constant value.

Option B (105) is correct because by using trigonometric identities and Vieta's formulas, the expression simplifies to 105.

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 D (4) is correct because there are four values of $\theta$ in the given interval that satisfy the equation.

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 C (Rational number) is correct because for the equation to have at least one solution, 'a' must be a rational number, ensuring that the arguments of sine and cosine are related in a way that allows both terms to reach their maximum values simultaneously.

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

Option B ($x = 2n\pi \, \pm \,{\cos ^{ - 1}}\,\left( {\frac{1}{3}} \right)$) is correct because it represents the general solution of the given trigonometric equation.

Option A ($n\\;\pi \\, + \\,{\left( { - 1} \right)^n}\frac{\pi }{4},\\,n \in Z$) is correct because it represents the general solution for $\theta$ derived from the given equation $\sec \theta \tan \theta = \sqrt 2$.

Option B ($n \pi+(-1)^{n+1}(\frac{3 \pi}{10})$) is correct because by solving the given trigonometric equation, we find the general solution for x.

Option A ($\tan \frac{\beta}{2} \tan 15^{\circ}$) is correct because it is derived by solving the given trigonometric equation for $\tan \frac{x}{2}$.

Option D ($\frac{1}{12}$) is correct because after simplifying the expression $f_4(x) - f_6(x)$, it equals $\frac{1}{12}$.

Option C, $\frac{\pi }{4}$, is correct because it is the only solution that satisfies the given equation $\sec \,x\,\cos \,5x + 1 = 0$ within the interval $0 < x < 2\pi$.

Option A ($2\;n\;\pi \\; \pm \\; \frac{\pi }{{3}},\\;n \in Z$) is correct because it satisfies the condition where the maximum value of the expression is 7.

Option D ($\sin \,2\theta$) is correct because by using the triple angle formula for sine and simplifying, we find that $x = \sin \,2\theta$.

Option D ($2 \mathrm{n} \pi+\frac{5 \pi}{6}$) is correct because it represents the general solution for θ that satisfies both $\sin \theta = \frac{1}{2}$ and $\cos \theta = -\frac{\sqrt{3}}{2}$, placing θ in the second quadrant.

Option C ($\cos \left( {\theta - \frac{\pi }{4}} \right) = \frac{1}{{2\sqrt 2 }}$) is correct because it is derived by using trigonometric identities to simplify the given equation.

Option C (5) is the correct answer because solving the trigonometric equation $\sin 3\theta = \sin \theta$ within the given interval yields five distinct solutions.

Option C ($\frac{{-\left|1-t^2\right|}}{{1+t^2}}$) is correct because, given $\sin \theta = \frac{{2t}}{{1+t^2}}$ and $\theta$ is in the second quadrant, $\cos \theta$ must be negative, leading to the expression $\frac{{-\left|1-t^2\right|}}{{1+t^2}}$.

Option A (2) is correct because solving the inequality $\frac{2x-1}{3} \geq \frac{3x-2}{4} - \frac{2-x}{5}$ results in $x \leq 2$, which corresponds to the interval $(-\infty, 2]$.

Option A ($\pm \sqrt{n \pi}, n \in{{0,1,2, \ldots}}$) is correct because it represents the set of all x values that satisfy the given functional equation after simplification and solving for x.

Option B ($I(n+2)$) is correct because by using the trigonometric identity $2 \cos a \cos b = \cos(a+b) + \cos(a-b)$, the expression simplifies to $I(n+2)$.