Embibe Experts Solutions for Chapter: Trigonometric Equations and Inequalities, Exercise 2: EXERCISE-2

The solution(s) of $4{\cos }^{2}x \sin x-2{\sin }^{2}x=3\sin x$ is/are

If $\left(\frac{1-a \sin x}{1+a \sin x}\right)\sqrt{\frac{1+2a \sin x}{1-2a \sin x}}=1$, where $a\in R$ then

The general solution of the following equation: $2\left(\sin x-\cos 2x\right)-\sin 2x\left(1+2\sin x\right)+2\cos x=0$ is/are

The value(s) of $\theta$, which satisfy the equation: $2{\cos }^{3}3\theta +3\cos 3\theta +4=3{\sin }^{2}3\theta$ is/are

Using four values of $\theta$ satisfying the equation $8{\cos }^4\theta +15{\cos }^2\theta -2=0$ in the interval $\left(0,4\pi \right)$, an arithmetic progression is formed, then:

The solution(s) of the equation $\cos 2x\sin 6x=\cos 3x\sin 5x$ in the interval $\left[0,\pi \right]$ is/are

The equation $4{\sin }^{2}x-2\left(\sqrt{3}+1\right)\sin x+\sqrt{3}=0$ has

If ${\cos }^{2}2x+2{\cos }^{2}x=1, x\in \left(-\pi ,\pi \right)$, then $x$ can take the values