What is the general solution for sin(1/2 x)+cos(x)=1 ?

The general solution for sin(1/2 x)+cos(x)=1 is x=4pin,x=2pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin

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sin(1/2 x)+cos(x)=1

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

x = 4 πn, x = 2 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn

Solution steps

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

Subtract

1

from both sides

sin (\frac{x}{2}) + cos (x) - 1 = 0

Let:

u = \frac{x}{2}

sin (u) + cos (2 u) - 1 = 0

Rewrite using trig identities

sin (u) - 2 sin^{2} (u) = 0

Solve by substitution

sin (u) = 0, sin (u) = \frac{1}{2}

sin (u) = 0 : u = 2 πn, u = π + 2 πn

sin (u) = \frac{1}{2} : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Combine all the solutions

u = 2 πn, u = π + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Substitute back

u = \frac{x}{2}

\frac{x}{2} = 2 πn : x = 4 πn

\frac{x}{2} = π + 2 πn : x = 2 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 = 4 πn, x = 2 π + 4 πn, x = \frac{π}{3} + 4 πn, x = \frac{5 π}{3} + 4 πn

cos (x) tan^{2} (x) = cos (x)

1 + 2 sin (\frac{x ^{2}}{2}) = 0

2 sin^{2} (x) + 7 cos (x) = 5

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ < 2 π

2 sin^{2} (x) = 3 sin (x)

What is the general solution for sin(1/2 x)+cos(x)=1 ?
The general solution for sin(1/2 x)+cos(x)=1 is x=4pin,x=2pi+4pin,x= pi/3+4pin,x=(5pi)/3+4pin