Embibe Experts Solutions for Chapter: Trigonometric Equations, Exercise 1: AP EAPCET 2020 (Sep-23 Shift-1)

Find the general solution of $3{\sin }^4\left(\theta \right)+{\cos }^4\left(\theta \right)=1$

If $cosec\theta +\cot \theta =\frac{1}{3},$ then $\theta$ lies in the

Assertion $\left(\mathrm{A}\right):$ If $\sqrt{4{\sin }^4\theta +{\sin }^{2}2\theta }+4{\cos }^2\left(\frac{\pi }{4}-\frac{\theta }{2}\right)=2,$ then $\theta$ lies in $3^{\mathrm{rd}}$ quadrant or $4^{\mathrm{th}}$ quadrant.
Reason $\left(\mathrm{R}\right): \sqrt{{\sin }^2\theta }=\sin \theta$

If $4\left(\sin 2x\sin 4x+{\sin }^{2}x\right)=3$, then $x=$

The number of solutions of the equation $4\cos 2\theta ·\cos 3\theta =\sec \theta ,$ when $0<\theta <\pi ,$ is

The value of $\text{'}x\text{'}$ satisfying the equation $3\mathrm{cosec}$ $x=4\sin x$ are

If $\theta \in [0,2\pi ]$ and $\cos 2\theta =\cos \theta +\sin \theta$ then the sum of all values of $\theta$ satisfying the equation is

If $\sin \left(\frac{\pi }{4}\cot \theta \right)=\cos \left(\frac{\pi }{4}\tan \theta \right)$, then $\theta =$