2-2cos^2(x/2)=2cos^2(x)

Solution

2 - 2 cos^{2} (\frac{x}{2}) = 2 cos^{2} (x)

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn, x = \frac{7 π}{3} + 4 πn

Solution steps

2 - 2 cos^{2} (\frac{x}{2}) = 2 cos^{2} (x)

Subtract

2 cos^{2} (x)

from both sides

2 - 2 cos^{2} (\frac{x}{2}) - 2 cos^{2} (x) = 0

Let:

u = \frac{x}{2}

2 - 2 cos^{2} (u) - 2 cos^{2} (2 u) = 0

Rewrite using trig identities

6 cos^{2} (u) - 8 cos^{4} (u) = 0

Solve by substitution

cos (u) = 0, cos (u) = \frac{3}{2}, cos (u) = - \frac{3}{2}

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = \frac{3}{2} : u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

cos (u) = - \frac{3}{2} : u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn, u = \frac{7 π}{6} + 2 πn

Substitute back

u = \frac{x}{2}

\frac{x}{2} = \frac{π}{2} + 2 πn : x = π + 4 πn

\frac{x}{2} = \frac{3 π}{2} + 2 πn : x = 3 π + 4 πn

\frac{x}{2} = \frac{π}{6} + 2 πn : x = \frac{π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{3} + 4 πn

\frac{x}{2} = \frac{5 π}{6} + 2 πn : x = \frac{5 π}{3} + 4 πn

\frac{x}{2} = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{3} + 4 πn

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn, x = \frac{7 π}{3} + 4 πn

sec^{2} (θ) - 2 = tan^{2} (θ)

cos (x) = sin (x) + 1

2 sin^{2} (θ) = 3 (1 - cos (- θ))

sec (x) - 2 = 0

2 sin^{4} (x) + sin^{2} (x) - 1 = 0