Solve 4(sin6⁡x+cos6⁡x)=4−3sin2⁡2x

Question

smallq9 · Accepted Answer

Firstly, work with the left side. Rewrite as a sum of two cubes:

4 (\sin^{6} x + \cos^{6} x) = 4 ((\sin^{2} x)^{3} + (\cos^{2} x)^{3})

Use the rule:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

4

(\sin^{x} + \cos^{6} x) = 4 (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + c o s^{4} x)

Use the Pythagorean identity:

\sin^{2} x + \cos^{2} x = 1

4 (\sin^{6} + \cos^{x}) = 4 (1) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4})

4 (\sin^{6} + \cos^{x}) = 4 (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4})

Add and substract

3 \sin^{2} x \cos^{2} x

to make a perfect square trinominal:

4 (\sin^{6} x + \cos^{6} x) = 4 (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x + 3 \sin^{2} x \cos^{2} x - 3 \sin^{2} x \cos^{2} x)

4 (\sin^{6} x + \cos^{6} x) = 4 (\sin^{4} x + 2 \sin^{2} x \cos^{2} x + \cos^{4} x - 3 \sin^{2} x \cos^{2} x)

4 (\sin^{6} x + \cos^{6} x) = 4 ((\sin^{2} x + \cos^{2} x)^{2} - 3 \sin^{2} x \cos^{2} x)

Use the Pythagorean identity:

\sin^{2} x + \cos^{2} x = 1

4 (\sin^{6} x + \cos^{6} x) = 4 ((1)^{2} - 3 \sin^{2} x \cos^{2} x)

4 (\sin^{6} x + \cos^{6} x) = 4 (1 - 3 \sin^{2} x \cos^{2} x)

4 (\sin^{6} x + \cos^{6} x) = 4 (1 - 3 (\sin^{2} x \cos^{2} x)^{2})

Use the double angle identity for sine:

2 x = 2 \sin x \cos x

4 (\sin^{6} x + \cos^{6}) = 4 (1 - 3 (\frac{\sin^{2} x}{2})^{2})

4 (\sin^{6} + \cos^{6}) = 4 (1 - 3 \times \frac{\sin^{2} 2 x}{4})

Solve 4(sin^{6}x+cos^{6}x)=4-3sin^{2} 2x