Question 885227
How do you solve
1-sinx=cos2x
***
for interval:[0,π/2]
{{{1-sinx=cos2x}}}
{{{1-sinx=cos^2(x)-sin^2(x)}}}
{{{1-sinx=1-sin^2(x)-sin^2(x)}}}
{{{1-sinx=1-2sin^2(x)}}}
{{{2sin^2(x)-sinx=0}}}
{{{sin(x)(2sinx-1)=0}}}
..
sinx=0
x=0
..
2sinx-1=0
sinx=1/2
x=π/6