What is the general solution for 9sin(x/2)+9cos(x)=0 ?

The general solution for 9sin(x/2)+9cos(x)=0 is x=(7pi)/3+4pin,x=(11pi)/3+4pin,x=pi+4pin

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9sin(x/2)+9cos(x)=0

9 sin (\frac{x}{2}) + 9 cos (x) = 0

x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = π + 4 πn

Solution steps

9 sin (\frac{x}{2}) + 9 cos (x) = 0

Let:

u = \frac{x}{2}

9 sin (u) + 9 cos (2 u) = 0

Rewrite using trig identities

(1 - 2 sin^{2} (u)) \cdot 9 + 9 sin (u) = 0

Solve by substitution

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

sin (u) = - \frac{1}{2} : u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

sin (u) = 1 : u = \frac{π}{2} + 2 πn

Combine all the solutions

u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn, u = \frac{π}{2} + 2 πn

Substitute back

u = \frac{x}{2}

\frac{x}{2} = \frac{7 π}{6} + 2 πn : x = \frac{7 π}{3} + 4 πn

\frac{x}{2} = \frac{11 π}{6} + 2 πn : x = \frac{11 π}{3} + 4 πn

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

x = \frac{7 π}{3} + 4 πn, x = \frac{11 π}{3} + 4 πn, x = π + 4 πn

- 5 cos (x) + 5 sin (x) = \frac{5 2}{2}

10 sin (θ) = 4 cos^{2} (θ) + 2

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

6 - 24 cos^{2} (t) = 0

sin^{2} (x) - sin (x) = 2, 0 \leq x \leq 2 π

What is the general solution for 9sin(x/2)+9cos(x)=0 ?
The general solution for 9sin(x/2)+9cos(x)=0 is x=(7pi)/3+4pin,x=(11pi)/3+4pin,x=pi+4pin