Question

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

Answer 1

Answer:

$\large\boxed{x=\dfrac{3\pi}{2}\ \vee\ x=\dfrac{\pi}{6}\ \vee\ x=\dfrac{5\pi}{6}}$

Step-by-step explanation:

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

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

$x=-\dfrac{\pi}{2}\notin[0,\ 2\pi)\\\\x=-\dfrac{\pi}{2}+2\pi=\dfrac{3\pi}{2}\in[0,\ 2\pi)\\\\x=-\dfrac{\pi}{2}+4\pi=\dfrac{7\pi}{2}\notin[0,\ 2\pi)\\\\x=\dfrac{\pi}{6}\in[0,\ 2\pi)\\\\x=\dfrac{\pi}{6}+2\pi=\dfrac{13\pi}{6}\notin[0,\ 2\pi)\\\\x=\dfrac{5\pi}{6}\in[0,\ 2\pi)\\\\x=\dfrac{5\pi}{6}+2\pi=\dfrac{17\pi}{6}\notin[0,\ 2\pi)$