9.5 Solving trigonometric equations  (Page 3/10)

Make sure mode is set to radians. To find $\theta ,$ use the inverse sine function. On most calculators, you will need to push the 2 ^ND button and then the SIN button to bring up the ${\sin }^{-1}$ function. What is shown on the screen is ${\sin }^{-1}(.$ The calculator is ready for the input within the parentheses. For this problem, we enter ${\sin }^{-1}\left(0.8\right),$ and press ENTER. Thus, to four decimals places,

The solution is

The angle measurement in degrees is

We can begin with some algebra.

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

Since $\frac{\pi }{2}\approx 1.57$ and $\pi \approx 3.14,$ 1.8235 is between these two numbers, thus $\theta \approx \text{1}\text{.8235}$ 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 [link] .

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

The solutions are $\theta \approx 1.8235\pm 2\pi k$ and $\theta \approx 4.4597\pm 2\pi k.$

We begin by using substitution and replacing cos $\theta$ with $x.$ It is not necessary to use substitution, but it may make the problem easier to solve visually. Let $\cos \theta =x.$ We have

The equation cannot be factored, so we will use the quadratic formula     $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$

Replace $x$ with $\cos \theta ,$ and solve.

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

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