Question

Solve the equation on the interval 0 ≤ θ < 2π.
please help
3 cot^2 θ - 4 csc θ = 1
a)pi/6 , 5pi/6
b)pi/6
c)7pi/6
d)7pi/6, 11pi/6

Answer 1

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[3\cot^2\theta-4\csc\theta=1\\
\frac{3\cos^2\theta}{\sin^2\theta}-\frac{4}{\sin\theta}=1\\
3\cos^2\theta-4\sin\theta=\sin^2\theta\\
3\left(1-\sin^2\theta\right)-4\sin\theta=\sin^2\theta\\
3-3\sin^2\theta-4\sin\theta-\sin^2\theta=0\\
3-4\sin\theta-4\sin^2\theta=0\\
(3+2\sin\theta)(1-2\sin\theta)=0\]
Solve the following:
\[\begin{cases}3+2\sin\theta=0\\
1-2\sin\theta=0\end{cases}\]
in the given interval.
\(\color{blue}{\text{End of Quote}}\)

Answer 2

okay thanks you

Answer 3

yw