What is the general solution for 3sin(ϕ)-cos(2ϕ)=1 ?

The general solution for 3sin(ϕ)-cos(2ϕ)=1 is ϕ= pi/6+2pin,ϕ=(5pi)/6+2pin

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3sin(ϕ)-cos(2ϕ)=1

3 sin (ϕ) - cos (2 ϕ) = 1

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

Solution steps

3 sin (ϕ) - cos (2 ϕ) = 1

Subtract

1

from both sides

3 sin (ϕ) - cos (2 ϕ) - 1 = 0

Rewrite using trig identities

- 2 + 2 sin^{2} (ϕ) + 3 sin (ϕ) = 0

Solve by substitution

sin (ϕ) = \frac{1}{2}, sin (ϕ) = - 2

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

sin (ϕ) = - 2 :

No Solution

Combine all the solutions

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

2 tan^{2} (θ) + 3 tan (θ) + 4 = 3

10 cos (x) = 8

2 sin^{2} (x) = 6 cos^{2} (x)

1 + cot^{2} (θ) = 1.33

x, r + s - 6 t = cos (2 x + y)

What is the general solution for 3sin(ϕ)-cos(2ϕ)=1 ?
The general solution for 3sin(ϕ)-cos(2ϕ)=1 is ϕ= pi/6+2pin,ϕ=(5pi)/6+2pin