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

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

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sin(x)=sqrt(2)cos(x/2)

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

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{2} + 4 πn, x = \frac{3 π}{2} + 4 πn

Solution steps

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

Subtract

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

from both sides

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

Let:

u = \frac{x}{2}

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

Rewrite using trig identities

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

Factor

- cos (u) 2 + 2 cos (u) sin (u) : cos (u) (- 2 + 2 sin (u))

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

Solving each part separately

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

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

- 2 + 2 sin (u) = 0 : u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 2 πn

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{π}{4} + 2 πn, u = \frac{3 π}{4} + 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}{4} + 2 πn : x = \frac{π}{2} + 4 πn

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

x = π + 4 πn, x = 3 π + 4 πn, x = \frac{π}{2} + 4 πn, x = \frac{3 π}{2} + 4 πn

2 sin (x) + \frac{1}{sin ( x )} = 3

cos (x - \frac{π}{4}) = \frac{3}{2}

sec (θ) = 3, sin (θ) < 0

cos (x) tan (x) - cos (x) = 0, 0 \leq x \leq 2 π

sin^{2} (2 θ) - 5 sin (2 θ) - 1 = 0

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