How do you solve 5sin^2x + 3cos^2x = 45sin2x+3cos2x=4?

1 Answer
Konstantinos Michailidis
Sep 15, 2015

The solutions are x=pi/4 ,x=(3pi)/4,x=(5pi)/4,x=(7pi)/4x=π4,x=3π4,x=5π4,x=7π4

Explanation:

We know that sin^2x+cos^2x=1sin2x+cos2x=1 hence cos^2x=1-sin^2xcos2x=1−sin2x

thus

5sin^2x+3(1-sin^2x)=4=>5sin^2x+3-3sin^2x=4=> 2sin^2x=1=>sinx=+-sqrt2/2=>x=pi/4 ,x=(3pi)/4,x=(5pi)/4,x=(7pi)/45sin2x+3(1−sin2x)=4⇒5sin2x+3−3sin2x=4⇒2sin2x=1⇒sinx=±√22⇒x=π4,x=3π4,x=5π4,x=7π4

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