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Solve for x and find all solutions in degrees from 0 to 360

20cos^2x+sin4x=13

 

I know you need to use the identity to turn cos^2x into 1-sin^2x but after I distribute I get lost. Please help me understand the question by writing out thought proccess thanks.

 Nov 24, 2019
edited by evanborot  Nov 24, 2019
edited by evanborot  Nov 24, 2019
edited by evanborot  Nov 24, 2019
 #1
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https://www.mathway.com/Algebra says there is no solutions that make this equation true for real numbers.

https://www.wolframalpha.com/input/?i=20cos%5E2%28x%29%2Bsin%284x%29%3D13 Well, wolfram doesn't have specific solution and some are complex no. 

 Nov 24, 2019
 #2
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I'll try to help

 

\(20(1-\sin^2x)+\sin4x=13\)

\(20-20\sin^2x+\sin4x=13\)

 

Bring all the sines to one side and numbers to the other.

 

\(20\sin^2x-\sin4x=7\)

 

Hopefully, from there you can find it.

 

You are very welcome!

:P

 Nov 24, 2019
 #3
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Can you confirm that the equation is correct ?

 

\(\displaystyle 20\cos^{2}x+\sin(4x)=13.\)

.
 Nov 25, 2019

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