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

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

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sin(x)=cos(x/2)

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

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

Solution steps

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

Subtract

cos (\frac{x}{2})

from both sides

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

Let:

u = \frac{x}{2}

sin (2 u) - cos (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

cos (\frac{π ( x - 7 )}{3}) = \frac{1}{2}

2 cos (2 x) + 8 cos (x) + 5 = 0

sin (x) \cdot tan (x) = 3 \cdot sin (x)

csc (x) = 2.6

2 sin^{2} (t) + 3 sin (t) + 1 = 0

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