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

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

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

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

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

Solution steps

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

1 + sin (u) - 2 sin^{2} (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

4 cos (2 θ) = 8 sin^{2} (θ)

cos (θ) = (\frac{5}{29})

sin (θ) = \frac{7.5}{6.5}

cos^{2} (x) - \frac{3}{2} = \frac{5}{2} cos (x)

θ, w = fd cos (θ)

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