Question

What is 1-2sin^2x=sinx
Please show steps and find all solutions in interval [0,2pi)

Answer 1

`\large\boxed{x=(3\pi)/(2)\ \vee\ x=(\pi)/(6)\ \vee\ x=(5\pi)/(6)}`

Explanation:

`1-2\sin^2x=\sin x\\\\\text{substitute}\ t=\sin x,\ t\in[-1,\ 1]\\\\1-2t^2=t\qquad\text{subtract t from both sides}\\\\-2t^2-t+1=0\qquad\text{change the signs}\\\\2t^2+t-1=0\\\\2t^2+2t-t-1=0\\\\2t(t+1)-1(t+1)=0\\\\(t+1)(2t-1)=0\iff t+1=0\ \vee\ 2t-1=0\\\\t+1=0\qquad\text{subtract 1 from both sides}\\\boxed{t=-1}\\\\2t-1=0\qquad\text{add 1 to both sides}\\2t=1\qquad\text{divide both sides by 2}\\\boxed{t=(1)/(2)}`

`\sin x=-1\to x=-(\pi)/(2)+2k\pi,\ k\in\mathbb{Z}\\\\\sin x=(1)/(2)\to x=(\pi)/(6)+2k\pi\ \vee\ x=(5\pi)/(6)+2k\pi,\ k\in\mathbb{Z}\\\\x\in[0,\ 2\pi)`

`x=-(\pi)/(2)\\otin[0,\ 2\pi)\\\\x=-(\pi)/(2)+2\pi=(3\pi)/(2)\in[0,\ 2\pi)\\\\x=-(\pi)/(2)+4\pi=(7\pi)/(2)\\otin[0,\ 2\pi)\\\\x=(\pi)/(6)\in[0,\ 2\pi)\\\\x=(\pi)/(6)+2\pi=(13\pi)/(6)\\otin[0,\ 2\pi)\\\\x=(5\pi)/(6)\in[0,\ 2\pi)\\\\x=(5\pi)/(6)+2\pi=(17\pi)/(6)\\otin[0,\ 2\pi)`