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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ/2)+cos(θ)=1

sin (\frac{θ}{2}) + cos (θ) = 1

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

Solution steps

sin (\frac{θ}{2}) + cos (θ) = 1

Subtract

1

from both sides

sin (\frac{θ}{2}) + cos (θ) - 1 = 0

Let:

u = \frac{θ}{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{θ}{2}

\frac{θ}{2} = 2 πn : θ = 4 πn

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

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

\frac{θ}{2} = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{3} + 4 πn

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

3 sin (\frac{π}{5} x) = 2

11 = 3 csc^{2} (x) + 7

cos (x) tan (x) = \frac{1}{2}

4 sin^{2} (x) = 1 - 4 cos (x)

- 100 sin (10 x) = 0

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