Question

find the general solution of sin 2theta=cos 3theta

Answer 1

Thew given equation is $\sin 2\theta=\cos 3\theta$. To find the general solution of the equation $\sin \theta = \sin \alpha,0\leq \alpha \leq 90^o$ is

$\theta = n \pi+ (-1)^n\alpha$

Now,

$\sin 2\theta=\cos 3\theta \\ \cos (90^o-2\theta)=\cos 3\theta \\ 90^o-2\theta=3\theta\\ 5\theta=90^o\\ 2\theta=36^o\\ \alpha =36^o$

Thus the general solution of the equation

$2 \theta = n \pi+ (-1)^n 36^o\\ \theta =\frac{ n \pi}{2}+ (-1)^n 18^o ,n=0,1,2,3,...$

Answer 2

Answer:

the upper answer is correct