What is the general solution for sin(θ)=cos(2θ+60) ?

The general solution for sin(θ)=cos(2θ+60) is θ=(2160n+180}{18},θ=-\frac{900+2160n)/6

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sin(θ)=cos(2θ+60)

sin (θ) = cos (2 θ + 6 0^{\circ})

θ = \frac{216 0 ^{\circ} n + 18 0 ^{\circ}}{18}, θ = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n}{6}

Solution steps

sin (θ) = cos (2 θ + 6 0^{\circ})

Rewrite using trig identities

sin (θ) = sin (9 0^{\circ} - (2 θ + 6 0^{\circ}))

Apply trig inverse properties

θ = 9 0^{\circ} - (2 θ + 6 0^{\circ}) + 36 0^{\circ} n, θ = 18 0^{\circ} - (9 0^{\circ} - (2 θ + 6 0^{\circ})) + 36 0^{\circ} n

θ = 9 0^{\circ} - (2 θ + 6 0^{\circ}) + 36 0^{\circ} n : θ = \frac{216 0 ^{\circ} n + 18 0 ^{\circ}}{18}

θ = 18 0^{\circ} - (9 0^{\circ} - (2 θ + 6 0^{\circ})) + 36 0^{\circ} n : θ = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n}{6}

θ = \frac{216 0 ^{\circ} n + 18 0 ^{\circ}}{18}, θ = - \frac{90 0 ^{\circ} + 216 0 ^{\circ} n}{6}

tan (θ) = \frac{- 12}{5}

\frac{1 + tan ( \frac{x}{2} )}{1 - tan ( \frac{x}{2} )} = 4.137131

cos (θ) = 0.51

csc (θ) = \frac{- 2 3}{3}

cos (x) = \frac{32}{50}

What is the general solution for sin(θ)=cos(2θ+60) ?
The general solution for sin(θ)=cos(2θ+60) is θ=(2160n+180}{18},θ=-\frac{900+2160n)/6