4cos(2t)=4sin^2(t)+1

Solution

4 cos (2 t) = 4 sin^{2} (t) + 1

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn, t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Solution steps

4 cos (2 t) = 4 sin^{2} (t) + 1

Subtract

4 sin^{2} (t) + 1

from both sides

4 cos (2 t) - 4 sin^{2} (t) - 1 = 0

Rewrite using trig identities

3 - 12 sin^{2} (t) = 0

Solve by substitution

sin (t) = \frac{1}{2}, sin (t) = - \frac{1}{2}

sin (t) = \frac{1}{2} : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn, t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

tan^{2} (θ) + 2 tan (θ) = - 1

cos (a) = \frac{2}{3}

- 1 = cos (π t)

2 tan^{2} (θ) = 6

cos^{2} (θ + \frac{π}{6}) = \frac{1}{2}