What is the general solution for sin(3θ+72)=cos(48),0<= θ<= 360 ?

The general solution for sin(3θ+72)=cos(48),0<= θ<= 360 is θ=22,θ=110,θ=142,θ=230,θ=262,θ=350

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sin(3θ+72)=cos(48),0<= θ<= 360

sin (3 θ + 7 2^{\circ}) = cos (4 8^{\circ}), 0^{\circ} \leq θ \leq 36 0^{\circ}

θ = 2 2^{\circ}, θ = 11 0^{\circ}, θ = 14 2^{\circ}, θ = 23 0^{\circ}, θ = 26 2^{\circ}, θ = 35 0^{\circ}

Solution steps

sin (3 θ + 7 2^{\circ}) = cos (4 8^{\circ}), 0^{\circ} \leq θ \leq 36 0^{\circ}

Rewrite using trig identities

sin (3 θ + 7 2^{\circ}) = sin (9 0^{\circ} - 4 8^{\circ})

Apply trig inverse properties

3 θ + 7 2^{\circ} = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n, 3 θ + 7 2^{\circ} = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n

3 θ + 7 2^{\circ} = 9 0^{\circ} - 4 8^{\circ} + 36 0^{\circ} n : θ = \frac{216 0 ^{\circ} n - 18 0 ^{\circ}}{18}

3 θ + 7 2^{\circ} = 18 0^{\circ} - (9 0^{\circ} - 4 8^{\circ}) + 36 0^{\circ} n : θ = \frac{198 0 ^{\circ} + 1080 0 ^{\circ} n}{90}

θ = \frac{216 0 ^{\circ} n - 18 0 ^{\circ}}{18}, θ = \frac{198 0 ^{\circ} + 1080 0 ^{\circ} n}{90}

Solutions for the range

0 \leq θ \leq 36 0^{\circ}

θ = 2 2^{\circ}, θ = 11 0^{\circ}, θ = 14 2^{\circ}, θ = 23 0^{\circ}, θ = 26 2^{\circ}, θ = 35 0^{\circ}

cot^{2} (y) + csc (y) - 5 = 0

2 cos (θ) = 1, 0 \leq θ < 2 π

cos^{2} (x) - tan (x) cos^{2} (x) = 0

3 tan (x) - 1 = 0, 0 \leq x \leq 2 π

6 + 12 sin (θ) = 0

What is the general solution for sin(3θ+72)=cos(48),0<= θ<= 360 ?
The general solution for sin(3θ+72)=cos(48),0<= θ<= 360 is θ=22,θ=110,θ=142,θ=230,θ=262,θ=350