Cos^2(x)+sin(x+π)=2sin(-x)?
Solve algebraically for all x in 0<or equal to x < or equal to 360
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Solve algebraically for all x in 0<or equal to x < or equal to 360
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this is cos² x + sin x = 0 ---> 1 - sin²x + sin x = 0...w² - w - 1 = 0 --->
w = [ 1 - √5 ] / 2---> x = arcsin [ (1-√5) / 2 ] = - 0.666239 + 2nπ or π + 0.666239 + 2nπ ;
- 38.172708°+ 360° OR 180° + 38.172708°