From the unit circle, we know that there will be two angles where [latex]\mathrm{cos}\left(\theta\right) =\frac{1}{2}[/latex] in one complete revolution, i.e. [latex]0\le\theta\le{2\pi}.[/latex] They will occur in the first and fourth quadrants. We recognize [latex]\frac{1}{2}[/latex] as a value from one of our special right triangles.  We can identify the acute angle as [latex]\frac{\pi}{3}.[/latex]  We can then use this as the reference angle to find the angle in the fourth quadrant by computing [latex]2\pi-\frac{\pi}{3}.[/latex]

These are the solutions in the interval [latex]\left[0,2\pi \right].[/latex] All possible solutions are given by  [latex]\frac{\pi }{3}\pm2k\pi[/latex] and [latex]\frac{5\pi }{3}\pm2k\pi[/latex] where [latex]k[/latex] is an integer.

Solving for all possible values of t means that solutions include angles beyond the period of [latex]2\pi .[/latex] From previous work with the unit circle, we know that there are two solutions in one revolution.  Since the value is positive, we know these solutions are in the first and second quadrants.  From our special right triangles, we know that the angle in the first quadrant is [latex]\frac{\pi }{6}[/latex] and using this as the reference angle, the solution in the second quadrant is [latex]\frac{5\pi }{6}.[/latex]  But the problem is asking for all possible values that solve the equation.

Therefore, the answer is  [latex]\frac{\pi }{6}\pm2\pi k[/latex] and [latex]\frac{5\pi }{6}\pm2\pi k[/latex]  where [latex]k[/latex] is an integer.

Use algebraic techniques to solve the equation.

We want all values of [latex]\theta[/latex] for which [latex]\mathrm{csc}\left(\theta\right) =-2[/latex] over the interval [latex]0\le \theta <4\pi.[/latex]

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

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

We can solve this equation using only algebra. Isolate the expression [latex]\mathrm{tan}\left(x\right)[/latex] on the left side of the equals sign.

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

We can begin with some algebra.

Check that the MODE is in radians. Now use the inverse cosine function.

Since [latex]\frac{\pi }{2}\approx 1.57[/latex] and [latex]\pi \approx 3.14,[/latex] we know that 1.8235 is between these two numbers, thus [latex]\theta \approx \text{1}\text{.8235}[/latex] is in quadrant II. Cosine is also negative in quadrant III. Note that a calculator will only return an angle in quadrants I or II for the cosine function, since that is the range of the inverse cosine. See Figure 1.

Figure 1

So, we also need to find the measure of the angle in quadrant III. In quadrant III, the reference angle is [latex]\theta^{\prime} \approx \pi -\text{1}\text{.8235}\approx \text{1}\text{.3181}\text{.}[/latex] The other solution in quadrant III is [latex]\pi +\text{1}\text{.3181}\approx \text{4}\text{.4597.}[/latex]

The solutions are [latex]1.8235\pm2\pi k[/latex] and [latex]4.4597\pm2\pi k.[/latex]

We begin by using substitution and replacing [latex]\mathrm{cos}\left(\theta\right)[/latex] with [latex]u.[/latex] It is not necessary to use substitution, but it may make the problem easier to solve visually. Let [latex]\mathrm{cos}\left(\theta\right) =u.[/latex] We have

The equation cannot be factored, so we will use the quadratic formula [latex]u=\frac{-b\pm\sqrt{{b}^{2}-4ac}}{2a},[/latex] where [latex]a=1,\text{ }b=3[/latex] and [latex]c=-1.[/latex]

Replace [latex]u[/latex] with [latex]\mathrm{cos}\left(\theta\right),[/latex] and solve. Thus,

Note that only the + sign is used. This is because we get an error when we solve [latex]\theta =\mathrm{cos}^{-1}\left(\frac{-3-\sqrt{13}}{2}\right)[/latex] on a calculator, since the domain of the inverse cosine function is [latex]\left[-1,1\right].[/latex] Therefore the solution is

This terminal side of the angle lies in quadrant I. Since cosine is also positive in quadrant IV, the second solution is

Now set each factor equal to zero and solve the two equations [latex]2\text{ }\mathrm{sin}\left(\theta\right) -3=0[/latex] and [latex]\mathrm{sin}\left(\theta\right) -1=0.[/latex]

For the first equation,

The first equation did not have any solution.  To solve the second equation,

We have three choices of expressions to substitute for the double-angle of cosine. As it is simpler to solve for one trigonometric function at a time, we will choose the double-angle identity involving only cosine:

[latex]\\[/latex]

Set the factors equal to zero and solve.

[latex]\begin{align*}2\mathrm{cos}\left(\theta\right)+1&=0&\mathrm{cos}\left(\theta\right)-1&=0\\\mathrm{cos}\left(\theta\right)&=-\frac{1}{2}&\mathrm{cos}\left(\theta\right)&=1\\\theta&=\frac{2\pi}{3}\pm2\pi k&\theta&=0\pm2\pi k\\\theta&=\frac{4\pi}{3}\pm2\pi k\end{align*}[/latex]