Question

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

Answer 1

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

Answer 2

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\[S=\left\{\frac{\pi}{2},\frac{\pi}{6},\frac{5\pi}{6}\right\}\]