Question

Use inverse trigonometric functions to find the solutions of the equation that are in the given interval, and approximate the solutions to four decimal places.
cos^2(x)+2cosx-2=0 [0, 2pi]

I'm having a little trouble with this. I believe I'm supposed to solve for cosine of x and then use the inverse cosine function to find x in radians?

Answer 1

So you have:
\[\cos^2(x)+2\cos(x)-2=0\]
This is actually a quadratic in disguise! Let \([u=\cos(x)]\). You have:
\[\eqalign{
&u^2+2u-2=0 \\
}\]
So then you can solve for \(u\) using the quad formula:
\[u=\frac{-2\pm\sqrt{4+4}}{2}=\frac{-2\pm\sqrt{8}}{2}=\frac{-2\pm2\sqrt{2}}{2}=-1\pm\sqrt{2}\]
So therefore, you can back substitute:
\[\cos(x)=-1\pm\sqrt{2}\]
Now, before I go any further, what grade is this for, or rather which class?

Answer 2

Its a trigonometry and analytic geometry course

Answer 3

In which grade?

Answer 4

Its a freshman college course

Answer 5

Its the same as a high school trig and precalc course essentially

Answer 6

Then you need to realize the intervals of \(\theta\) where \(\cos(x)\) is positive and negative:

So then you know that if \(\cos(x)=+ve\), then it must be in the first or last quadrant. Also, if \(\cos(x)=-ve\), then it must be in the second or third quadrant.