Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2sin^2 X -3sinX+1=0

Question

anonymous · Answer

$$2\sin^2x-3\sin x+1=(2\sin x-1)(\sin x-1)$$

anonymous · Answer

replace sin(x) by y$$2sin^2(x)-3sin(x)+1$$$$2y^2-3y+1$$therefore$$y_1=1~~and~~y_2=\frac{1}{2}$$Now we have to replace y by sin(x) to find the value of X
if y=1$$sin(x)=1$$$$x=\frac{\pi}{2}=90^o$$if y=1$$sin(x)=\frac{1}{2}$$$$x=\frac{\pi}{6}=30^o~~or~~x=\frac{5\pi}{6}=150^o$$The answer is$$S=\left\{\frac{\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}\right\}$$