What is the general solution for 8sin(x/2)+8cos(x)=0 ?

The general solution for 8sin(x/2)+8cos(x)=0 is x=(7pi)/3+4pin,x=(11pi)/3+4pin,x=pi+4pin

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8sin(x/2)+8cos(x)=0

8 sin (\frac{x}{2}) + 8 cos (x) = 0

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

Solution steps

8 sin (\frac{x}{2}) + 8 cos (x) = 0

Let:

u = \frac{x}{2}

8 sin (u) + 8 cos (2 u) = 0

Rewrite using trig identities

(1 - 2 sin^{2} (u)) \cdot 8 + 8 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

tan (\frac{θ}{2} + \frac{π}{6}) = 1

sec (x) = - 2, 0 < x < 2 π

1 - cos (2 x) = 0

sin (2 x) cos (x) = cos (x)

- sin (2 x) sin (x) + cos (2 x) cos (x) = cos (x)

What is the general solution for 8sin(x/2)+8cos(x)=0 ?
The general solution for 8sin(x/2)+8cos(x)=0 is x=(7pi)/3+4pin,x=(11pi)/3+4pin,x=pi+4pin