What is the general solution for 1+cos(2θ)=3sin(θ) ?

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

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1+cos(2θ)=3sin(θ)

1 + cos (2 θ) = 3 sin (θ)

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

Solution steps

1 + cos (2 θ) = 3 sin (θ)

Subtract

3 sin (θ)

from both sides

1 + cos (2 θ) - 3 sin (θ) = 0

Rewrite using trig identities

2 - 2 sin^{2} (θ) - 3 sin (θ) = 0

Solve by substitution

sin (θ) = - 2, sin (θ) = \frac{1}{2}

sin (θ) = - 2 :

No Solution

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

Combine all the solutions

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

2 cos (x) sin (x) - 3 sin (x) = 0

cos (θ) - sin (θ) = cot (θ) cos (θ)

- 2 tan (x) + 1 + 2 tan^{2} (x) = tan^{2} (x)

\frac{sin ( a )}{50} = \frac{sin ( 2 5 ^{\circ} )}{35}

sin (x + \frac{π}{4}) = \pm 1

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