What is the general solution for 4sin(θ)=4cos(2θ) ?

The general solution for 4sin(θ)=4cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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4sin(θ)=4cos(2θ)

4 sin (θ) = 4 cos (2 θ)

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solution steps

4 sin (θ) = 4 cos (2 θ)

Subtract

4 cos (2 θ)

from both sides

4 sin (θ) - 4 cos (2 θ) = 0

Rewrite using trig identities

- (1 - 2 sin^{2} (θ)) \cdot 4 + 4 sin (θ) = 0

Solve by substitution

sin (θ) = \frac{1}{2}, sin (θ) = - 1

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

sin (θ) = - 1 : θ = \frac{3 π}{2} + 2 πn

Combine all the solutions

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn, θ = \frac{3 π}{2} + 2 πn

tan (5 θ + 30) = 1

885 cos (θ) - 70 = cos (25 0^{\circ})

tan (x) + 5 = 0, x \leq 0 < 2 π

\frac{31.5}{sin ( 57.2 2 ^{\circ} )} = \frac{24.4}{sin ( x )}

23 cos (θ - \frac{π}{3}) = 3, 0 \leq θ \leq 2 π

What is the general solution for 4sin(θ)=4cos(2θ) ?
The general solution for 4sin(θ)=4cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin