9.5 Solving trigonometric equations  (Page 2/10)

Solving for all possible values of t means that solutions include angles beyond the period of $2\pi .$ From [link] , we can see that the solutions are $t=\frac{\pi }{6}$ and $t=\frac{5\pi }{6}.$ But the problem is asking for all possible values that solve the equation. Therefore, the answer is

where $k$ is an integer.

Use algebraic techniques to solve the equation.

As this problem is not easily factored, we will solve using the square root property. First, we use algebra to isolate $\sin \theta .$ Then we will find the angles.

We want all values of $\theta$ for which $\csc \theta =-2$ over the interval $0\le \theta <4\pi .$

Recall that the tangent function has a period of $\pi .$ On the interval $\left[0,\pi \right),$ and at the angle of $\frac{\pi }{4},$ the tangent has a value of 1. However, the angle we want is $\left(\theta -\frac{\pi }{2}\right).$ Thus, if $\tan \left(\frac{\pi }{4}\right)=1,$ then

Over the interval $\left[0,2\pi \right),$ we have two solutions:

We can solve this equation using only algebra. Isolate the expression $\tan x$ on the left side of the equals sign.

There are two angles on the unit circle that have a tangent value of $\mathrm{-1}\text{:}\theta =\frac{3\pi }{4}$ and $\theta =\frac{7\pi }{4}.$