What is the general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) ?

The general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) is a=2pin,a=pi+2pin

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1-tan^4(a)cos^4(a)=1-2sin^2(a)

1 - tan^{4} (a) cos^{4} (a) = 1 - 2 sin^{2} (a)

a = 2 πn, a = π + 2 πn

Solution steps

1 - tan^{4} (a) cos^{4} (a) = 1 - 2 sin^{2} (a)

Subtract

1 - 2 sin^{2} (a)

from both sides

2 sin^{2} (a) - tan^{4} (a) cos^{4} (a) = 0

Factor

2 sin^{2} (a) - tan^{4} (a) cos^{4} (a) : (2 sin (a) + tan^{2} (a) cos^{2} (a)) (2 sin (a) - tan^{2} (a) cos^{2} (a))

(2 sin (a) + tan^{2} (a) cos^{2} (a)) (2 sin (a) - tan^{2} (a) cos^{2} (a)) = 0

Solving each part separately

2 sin (a) + tan^{2} (a) cos^{2} (a) = 0 or 2 sin (a) - tan^{2} (a) cos^{2} (a) = 0

2 sin (a) + tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

2 sin (a) - tan^{2} (a) cos^{2} (a) = 0 : a = 2 πn, a = π + 2 πn

Combine all the solutions

a = 2 πn, a = π + 2 πn

cos (0.5 x) = 0.5, - π \leq x \leq π

0.5 cos (x) = cos (x)

tan (a) = - \frac{1}{2}

sin (2 x + 3 0^{\circ}) = - \frac{3}{2}

8 tan^{2} (θ) - 1 = 7 tan (θ)

What is the general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) ?
The general solution for 1-tan^4(a)cos^4(a)=1-2sin^2(a) is a=2pin,a=pi+2pin