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

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

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cos(x/2)=sin(x)

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

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

Solution steps

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

Subtract

sin (x)

from both sides

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Factor

cos (u) - 2 cos (u) sin (u) : - cos (u) (2 sin (u) - 1)

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

Solving each part separately

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

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

2 sin (u) - 1 = 0 : u = \frac{π}{6} + 2 πn, u = \frac{5 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{5 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{5 π}{6} + 2 πn : x = \frac{5 π}{3} + 4 πn

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

4 sin^{2} (x) - 5 sin (x) + 1 = 0

tan (x) = - 0.6293

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

- sin (2 x) - sin (x) = 0

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

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