What is the general solution for 7sin(θ)=7cos(2θ) ?

The general solution for 7sin(θ)=7cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin

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7sin(θ)=7cos(2θ)

7 sin (θ) = 7 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

7 sin (θ) = 7 cos (2 θ)

Subtract

7 cos (2 θ)

from both sides

7 sin (θ) - 7 cos (2 θ) = 0

Rewrite using trig identities

- (1 - 2 sin^{2} (θ)) \cdot 7 + 7 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

sin^{2} (x) - cos^{2} (x) - cos (x) - 1 = 0

t, y^{(4)} - 81 y = sin (t)

tan (\frac{θ}{2} + \frac{π}{6}) = - 1, 0 \leq θ < 2 π

3 sec^{2} (x) - 3 = 0

tan (θ) = 19, 0 \leq θ < 2 π

What is the general solution for 7sin(θ)=7cos(2θ) ?
The general solution for 7sin(θ)=7cos(2θ) is θ= pi/6+2pin,θ=(5pi)/6+2pin,θ=(3pi)/2+2pin