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

The general solution for sin(1/2 x)=cos(x) is x= pi/3+4pin,x=(5pi)/3+4pin,x=3pi+4pin

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

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

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

Solution steps

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

Subtract

cos (x)

from both sides

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Solve by substitution

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

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

sin (u) = - 1 : u = \frac{3 π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{x}{2}

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

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

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

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

tan (2 x - \frac{π}{4}) = - 1

2 sin^{2} (A) = 5 sin (A) - 2

cot (θ) = \frac{12}{5}

θ, cos (θ) = - \frac{1}{2}

- 3 sin (b) - 2 = sin (b) - 2

What is the general solution for sin(1/2 x)=cos(x) ?
The general solution for sin(1/2 x)=cos(x) is x= pi/3+4pin,x=(5pi)/3+4pin,x=3pi+4pin