What is the general solution for cos(x/2)=cos(x)+1 ?

The general solution for cos(x/2)=cos(x)+1 is x=pi+4pin,x=3pi+4pin,x=(2pi)/3+4pin,x=(10pi)/3+4pin

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cos(x/2)=cos(x)+1

cos (\frac{x}{2}) = cos (x) + 1

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

Solution steps

cos (\frac{x}{2}) = cos (x) + 1

Subtract

cos (x) + 1

from both sides

cos (\frac{x}{2}) - cos (x) - 1 = 0

Let:

u = \frac{x}{2}

cos (u) - cos (2 u) - 1 = 0

Rewrite using trig identities

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

Solve by substitution

cos (u) = 0, cos (u) = \frac{1}{2}

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

cos (u) = \frac{1}{2} : u = \frac{π}{3} + 2 πn, u = \frac{5 π}{3} + 2 πn

Combine all the solutions

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

\frac{x}{2} = \frac{5 π}{3} + 2 πn : x = \frac{10 π}{3} + 4 πn

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

d = \frac{sin ^{4} ( x ) - cos ^{2} ( x ) + 5}{4 \cdot cos ^{2} ( x )}

2 + cos^{2} (x) = 5 sin (x)

tan^{3} (3 x) - 2 sin^{3} (3 x) = 0

cot^{5} (x) = \frac{- 1}{( 3 )}

2 cos^{4} (x) cos (x) - cos^{5} (x) = 1

What is the general solution for cos(x/2)=cos(x)+1 ?
The general solution for cos(x/2)=cos(x)+1 is x=pi+4pin,x=3pi+4pin,x=(2pi)/3+4pin,x=(10pi)/3+4pin