What is the general solution for cos(x/2)-sin(x)=0 ?

The general solution for cos(x/2)-sin(x)=0 is x=pi+4pin,x=3pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin

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

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

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) = 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

2 tan^{2} (θ) + tan (θ) = 0

2 cos^{2} (θ) + sin (θ) = 1

tan (2 x) - 3 = 0

0 = cos (2 x)

cos (5 x) = 0

What is the general solution for cos(x/2)-sin(x)=0 ?
The general solution for cos(x/2)-sin(x)=0 is x=pi+4pin,x=3pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin