Solve (sin x + 1)(2 sin^2 x - 3 sin x - 2) on the interval [0, 2pi)

Question

.sam. · Answer

\[\begin{array}{l} (\sin (x)+1) \left(2 \sin ^2(x)-3 \sin (x)-2\right)=0 \ \text{Split into }2\text{ equations:} \ \sin (x)+1=0\text{ or }2 \sin ^2(x)-3 \sin (x)-2=0 \ \text{Subtract }1\text{ from both sides:} \ \sin (x)=-1\text{ or }2 \sin ^2(x)-3 \sin (x)-2=0 \ \text{Take the inverse sine of both sides:} \ x=-\frac{\pi }{2}\text{ or }2 \sin ^2(x)-3 \sin (x)-2=0 \ \text{Substitute }u=\sin (x): \ x=-\frac{\pi }{2}\text{ or }2 u^2-3 u-2=0 \ \text{Solve the quadratic equation by completing the \square:}\text{$\backslash $n$\backslash $n}\text{Divide both sides by }2: \ x=-\frac{\pi }{2}\text{ or }u^2-\frac{3 u}{2}-1=0 \ \text{Add }1\text{ \to both sides:} \ x=-\frac{\pi }{2}\text{ or }u^2-\frac{3 u}{2}=1 \ \text{Add }\frac{9}{16}\text{ \to both sides:} \ x=-\frac{\pi }{2}\text{ or }u^2-\frac{3 u}{2}+\frac{9}{16}=\frac{25}{16} \ \text{Factor the \left hand side:} \ x=-\frac{\pi }{2}\text{ or }\left(u-\frac{3}{4}\right)^2=\frac{25}{16} \ \text{Take the \square \root of both sides:} \ x=-\frac{\pi }{2}\text{ or }\left|u-\frac{3}{4}\right|=\frac{5}{4} \ \text{Eliminate the absolute value:} \ x=-\frac{\pi }{2}\text{ or }u-\frac{3}{4}=-\frac{5}{4}\text{ or }u-\frac{3}{4}=\frac{5}{4} \ \text{Add }\frac{3}{4}\text{ \to both sides:} \ x=-\frac{\pi }{2}\text{ or }u=-\frac{1}{2}\text{ or }u-\frac{3}{4}=\frac{5}{4} \ \text{Substitute back for }u=\sin (x): \ x=-\frac{\pi }{2}\text{ or }\sin (x)=-\frac{1}{2}\text{ or }u-\frac{3}{4}=\frac{5}{4} \ \text{Take the inverse sine of both sides:} \ x=-\frac{\pi }{2}\text{ or }x=-\frac{\pi }{6}\text{ or }u-\frac{3}{4}=\frac{5}{4} \ \text{Add }\frac{3}{4}\text{ \to both sides:} \ x=-\frac{\pi }{2}\text{ or }x=-\frac{\pi }{6}\text{ or }u=2 \ \text{Substitute back for }u=\sin (x): \ x=-\frac{\pi }{2}\text{ or }x=-\frac{\pi }{6}\text{ or }\sin (x)=2 \ \text{Take the inverse sine of both sides:} \ x=-\frac{\pi }{2}\text{ or }x=-\frac{\pi }{6}\text{ or }x=\sin ^{-1}(2) \\end{array}\]