Can someone give me a clue, how can i solve this? I want something like: $x=pi/4+kpi$ ---> Solve this: $2sin^2(x)-sin^2(2x)=cos^2(2x)$

Question

$2sin^2(x)-sin^2(2x)=cos^2(2x)$

It should be:

$x=pi/4+kpi$ AND $x=(3pi)/4+kpi$

3536 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-06T16:01:44

See below.

You can write also

$2sin^2x=sin^2(2x)+cos^2(2x)=1$

so

$sin^2x=1/2->sinx=pm sqrt(2)/2$

so

$x = pm pi/4+2kpi$ for $k in ZZ$

Answer 2 · 2017-04-06T16:08:21

Given: $2sin^2(x)-sin^2(2x)=cos^2(2x)$

Add $sin^2(2x)$ to both sides:

$2sin^2(x)=sin^2(2x)+cos^2(2x)$

Substitute 1 for $sin^2(2x)+cos^2(2x)$ :

$2sin^2(x)=1$

$sin^2(x)=1/2$

$sin(x) = +-sqrt(2)/2$

$x = sin^-1(sqrt(2)/2) and x = sin^-1(-sqrt(2)/2)$

$x = pi/4+kpi and x = (3pi)/4+ kpi; k in ZZ$