The solution of equation cos^2theta - 2cos theta = 4sin theta - sin2theta where theta belongs to 0 to pi is

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θ

3 - 2 c o s θ - 4 s i n θ - c o s 2 θ + s i n 2 θ = 0

sin 7 θ = sin θ + sin 3 θ f o r θ \in (0, π)

sin2θ + cos2θ	=	1.

Therefore is ROOT of sin2theta + ROOT of cos2theta must be equal to ROOT of one ?

Basically is sin ROOT theta +cos ROOT theta = 1

(1 - 2 c o s θ)^{2} + (t a n θ + \sqrt{3})^{2} = 0

[0, 2 π]

the question given below is of matrix :-

If A=[cos2theta sin2theta]

[-sin2theta cos2theta] find A^2

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