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

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

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sin(2θ)=3cos(θ)

sin (2 θ) = 3 cos (θ)

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (2 θ) = 3 cos (θ)

Subtract

3 cos (θ)

from both sides

sin (2 θ) - 3 cos (θ) = 0

Rewrite using trig identities

sin (2 θ) - 3 cos (θ)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 2 sin (θ) cos (θ) - 3 cos (θ)

- 3 cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

- 3 cos (θ) + 2 cos (θ) sin (θ) : cos (θ) (2 sin (θ) - 3)

- 3 cos (θ) + 2 cos (θ) sin (θ)

Factor out common term

cos (θ)

= cos (θ) (- 3 + 2 sin (θ))

cos (θ) (2 sin (θ) - 3) = 0

Solving each part separately

cos (θ) = 0 or 2 sin (θ) - 3 = 0

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

2 sin (θ) - 3 = 0 :

No Solution

2 sin (θ) - 3 = 0

Move

3

to the right side

2 sin (θ) - 3 = 0

Add

3

to both sides

2 sin (θ) - 3 + 3 = 0 + 3

Simplify

2 sin (θ) = 3

2 sin (θ) = 3

Divide both sides by

2

2 sin (θ) = 3

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{3}{2}

Simplify

sin (θ) = \frac{3}{2}

sin (θ) = \frac{3}{2}

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

x, tan (x) + ∣ tan (x) ∣ = 1

0 = 2 sin^{2} (θ) + 3 sin (θ)

y, x = arctan (x + y)

2 sin (2 x) = 3 tan (x)

tan (φ) = \frac{1}{3}

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