What is the general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 ?

The general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 is θ=90,θ=0.89112…,θ=180-0.89112…

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-9cos^2(θ)+9=16sin(θ)-7,0<= θ<360

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

θ = 9 0^{\circ}, θ = 0.89112 \dots, θ = 18 0^{\circ} - 0.89112 \dots

Solution steps

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

Subtract

16 sin (θ) - 7

from both sides

- 9 cos^{2} (θ) - 16 sin (θ) + 16 = 0

Rewrite using trig identities

7 - 16 sin (θ) + 9 sin^{2} (θ) = 0

Solve by substitution

sin (θ) = 1, sin (θ) = \frac{7}{9}

sin (θ) = 1, 0 \leq θ < 36 0^{\circ} : θ = 9 0^{\circ}

sin (θ) = \frac{7}{9}, 0 \leq θ < 36 0^{\circ} : θ = arcsin (\frac{7}{9}), θ = 18 0^{\circ} - arcsin (\frac{7}{9})

Combine all the solutions

θ = 9 0^{\circ}, θ = arcsin (\frac{7}{9}), θ = 18 0^{\circ} - arcsin (\frac{7}{9})

Show solutions in decimal form

θ = 9 0^{\circ}, θ = 0.89112 \dots, θ = 18 0^{\circ} - 0.89112 \dots

\frac{cos ( x )}{sin ^{2} ( x )} = \frac{2}{3}

7 cos (b) - 13 = 0

cos (θ) = \frac{6}{3.92739}

sin (x) = - \frac{1}{3}, 0 \leq x \leq 2 π

3 csc (x) = 2 tan (x)

What is the general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 ?
The general solution for -9cos^2(θ)+9=16sin(θ)-7,0<= θ<360 is θ=90,θ=0.89112…,θ=180-0.89112…