Question

2sin^2 2x=2

In the domain [0,2pi)

What can x equal ( in radians).

Answer 1

bear in mind that, when it comes to trigonometric functions, the location of the exponent can be a bit misleading, however recall that sin²(θ) is really [ sin( θ )]²,



`\bf 2sin^2(2x)=2\implies sin^2(2x)=\cfrac{2}{2} \\\\\\ sin^2(2x)=1\implies [sin(2x)]^2=1\implies sin(2x)=\pm√(1) \\\\\\ sin(2x)=\pm 1\implies sin^(-1)[sin(2x)]=sin^(-1)(\pm 1)`


`\bf \measuredangle 2x=sin^(-1)(\pm 1)\implies \measuredangle 2x= \begin{cases} (\pi )/(2)\\\\ (3\pi )/(2) \end{cases}\\\\ -------------------------------\\\\ \measuredangle 2x=\cfrac{\pi }{2}\implies \measuredangle x=\cfrac{\pi }{4}\qquad \qquad \qquad \qquad \measuredangle 2x=\cfrac{3\pi }{2}\implies \measuredangle x=\cfrac{3\pi }{4}`