What is the general solution for cos(θ)+3sin(θ/2)-2=0 ?

The general solution for cos(θ)+3sin(θ/2)-2=0 is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=pi+4pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(θ)+3sin(θ/2)-2=0

cos (θ) + 3 sin (\frac{θ}{2}) - 2 = 0

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

Solution steps

cos (θ) + 3 sin (\frac{θ}{2}) - 2 = 0

Let:

u = \frac{θ}{2}

cos (2 u) + 3 sin (u) - 2 = 0

Rewrite using trig identities

- 1 - 2 sin^{2} (u) + 3 sin (u) = 0

Solve by substitution

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

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

sin (u) = 1 : u = \frac{π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

\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

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

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

2 sec (θ) = 6

sin (2 x + 20) - 1 = \frac{- 1}{2}

x, sin (x + \frac{π}{2}) - cos^{2} (x) = 0

0 = 4 + 3 cos (θ)

cos (θ) = \frac{6}{8}

What is the general solution for cos(θ)+3sin(θ/2)-2=0 ?
The general solution for cos(θ)+3sin(θ/2)-2=0 is θ= pi/3+4pin,θ=(5pi)/3+4pin,θ=pi+4pin