What is the general solution for sin(θ)=sqrt(3)cos(θ),0<= θ<2pi ?

The general solution for sin(θ)=sqrt(3)cos(θ),0<= θ<2pi is θ= pi/3 ,θ=(4pi)/3

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sin(θ)=sqrt(3)cos(θ),0<= θ<2pi

sin (θ) = 3 cos (θ), 0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{4 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 24 0^{\circ}

Solution steps

sin (θ) = 3 cos (θ), 0 \leq θ < 2 π

Rewrite using trig identities

sin (θ) = 3 cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{sin ( θ )}{cos ( θ )} = \frac{3 cos ( θ )}{cos ( θ )}

Simplify

\frac{sin ( θ )}{cos ( θ )} = 3

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

tan (θ) = 3

tan (θ) = 3

General solutions for

tan (θ) = 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{4 π}{3}

cos (2 x) - \frac{1}{3} cos (x) = 0

3 = 23 sin (x)

8 cos (6 x) = 2

sin (x) = \frac{- ( 6 )}{( 7 ) cos ( x )}

5^{2} = 8^{2} - (11.85)^{2} - 2 (8) (11.85) (cos (A))

What is the general solution for sin(θ)=sqrt(3)cos(θ),0<= θ<2pi ?
The general solution for sin(θ)=sqrt(3)cos(θ),0<= θ<2pi is θ= pi/3 ,θ=(4pi)/3