Answer to Question #223092 in Trigonometry for Vanessam

Question

Solve the following equations

2sin2x-3cosx = 0

sinx + cosx =0

Answer 1

$2sin2x-3cosx\:=\:0\:$

$-3\cos \:\left(x\right)+4\cos \left(x\right)\sin \left(x\right)=0$

$\cos \left(x\right)\left(4\sin \left(x\right)-3\right)=0$

$\cos \left(x\right)=0\quad \:\mathrm{or}\quad \:\:4\sin \left(x\right)-3=0$

$x=\frac{\pi \:}{2}+2\pi \:n,\:x=\frac{3\pi \:}{2}+2\pi \:n,\:x=\arcsin \:\left(\frac{3}{4}\right)+2\pi \:n,\:x=\pi \:-\arcsin \:\left(\frac{3}{4}\right)+2\pi \:n$

$x=\frac{\pi }{2}+2\pi n,\:x=\frac{3\pi }{2}+2\pi n,\:x=0.84806\dots +2\pi n,\:x=\pi -0.84806\dots +2\pi n$

$sinx\:+\:cosx\:=0$

$\tan \left(x\right)+1=0$

$\tan \left(x\right)+1-1=0-1$

$\tan \left(x\right)=-1$

$x=\frac{3\pi \:}{4}+\pi \:n$