Question

Find all solutions of the equation 2sin2x−cosx=1 in the interval [0,2π), what is x?

Answer 1

This is a little it hard... I am wondering if others have a better way to solve it.

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2sin2x−cosx=1 

4sinxcosx = 1 + cosx 

16sin^2xcos^2x = 1 + 2cosx + cos^2x 

16(1 - cos^2x)cos^2x = 1 + 2cosx + cos^2x 

16cos^2x - 16cos^4x = 1 + 2cosx + cos^2x 

16cos^4x - 15cos^2x + 2cosx + 1 = 0 

let cosx = Y 

16Y^4 - 15Y^2 + 2Y + 1 = 0 

(Y + 1)(16Y^3 - 16Y^2 + Y + 1) = 0 


there are four values for Y, -1, -0.2, 0.367, 0.836. 

Then you solve for Y. 

For example, Y = -1 means cosx = -1, x = π. 

I will leave the rests to you. (Sorry, this seems to be an ugly way...)