Question

3sinx = 2cos^2x

solve.

Answer 1

Graph them on your calculator my friend and find the intersecting points ;)

Answer 2

no i have to solve them without a calculator

Answer 3

Just clarifying that 2cos^2x is\[2\cos^2x\]

Answer 4

\[3\sin x=2\cos ^{2}x\]

\[\cos ^{2}x+\sin ^{2}x=1\]

\[\cos ^{2}x=1-\sin ^{2}x\]
\[3\sin x=2(1-\sin ^{2}x)\]

\[2\sin ^{2}x+3sinx-2=0\]
(sinx+2)(2sinx-1)=0

1)sinx+2=0 => sinx=-2 undefined

2)2sinx-1=0 =>sinx=1/2 =>x=30 or pi/6

Answer 5

\[-1\le sinx \le1\]