What is the general solution for sin(3θ+72)=cos(48) ?

The general solution for sin(3θ+72)=cos(48) is θ=(10800n+1260-2160)/(90),θ=(4140+10800n-2160)/(90)

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sin(3θ+72)=cos(48)

sin (3 θ + 72) = cos (4 8^{\circ})

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}, θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

Solution steps

sin (3 θ + 72) = cos (4 8^{\circ})

Rewrite using trig identities

sin (3 θ + 72) = sin (9 0^{\circ} - 4 8^{\circ})

Apply trig inverse properties

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n, 3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

3 θ + 72 = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n : θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}

3 θ + 72 = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n : θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

θ = \frac{1080 0 ^{\circ} n + 126 0 ^{\circ} - 2160}{90}, θ = \frac{414 0 ^{\circ} + 1080 0 ^{\circ} n - 2160}{90}

tan (x) = \frac{1}{5}

sec (x) = 2 cos (x)

sec (5 \frac{θ}{4}) = 2, 0 < θ < 2 π

2 sin (\frac{x}{2}) = 1

sin^{2} (x) - 3 cos^{2} (x) = 0

What is the general solution for sin(3θ+72)=cos(48) ?
The general solution for sin(3θ+72)=cos(48) is θ=(10800n+1260-2160)/(90),θ=(4140+10800n-2160)/(90)