What is the general solution for sin(θ)=cos(2θ+54) ?

The general solution for sin(θ)=cos(2θ+54) is θ=(1800n+180}{15},θ=-\frac{720+1800n)/5

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sin(θ)=cos(2θ+54)

sin (θ) = cos (2 θ + 5 4^{\circ})

θ = \frac{180 0 ^{\circ} n + 18 0 ^{\circ}}{15}, θ = - \frac{72 0 ^{\circ} + 180 0 ^{\circ} n}{5}

Solution steps

sin (θ) = cos (2 θ + 5 4^{\circ})

Rewrite using trig identities

sin (θ) = sin (9 0^{\circ} - (2 θ + 5 4^{\circ}))

Apply trig inverse properties

θ = 9 0^{\circ} - (2 θ + 5 4^{\circ}) + 36 0^{\circ} n, θ = 18 0^{\circ} - (9 0^{\circ} - (2 θ + 5 4^{\circ})) + 36 0^{\circ} n

θ = 9 0^{\circ} - (2 θ + 5 4^{\circ}) + 36 0^{\circ} n : θ = \frac{180 0 ^{\circ} n + 18 0 ^{\circ}}{15}

θ = 18 0^{\circ} - (9 0^{\circ} - (2 θ + 5 4^{\circ})) + 36 0^{\circ} n : θ = - \frac{72 0 ^{\circ} + 180 0 ^{\circ} n}{5}

θ = \frac{180 0 ^{\circ} n + 18 0 ^{\circ}}{15}, θ = - \frac{72 0 ^{\circ} + 180 0 ^{\circ} n}{5}

t, y = cos (t)

x, r^{2} = 9 cos (2 x)

csc (a) = \frac{13}{12}

(cos (θ) + 1) (tan (θ) + 1) = 0

(tan^{2} (3 x)) (csc (3 x)) + 1 = - 1 \cdot sec^{2} (3 x)

What is the general solution for sin(θ)=cos(2θ+54) ?
The general solution for sin(θ)=cos(2θ+54) is θ=(1800n+180}{15},θ=-\frac{720+1800n)/5