- Thread starter
- #1

#### Petrus

##### Well-known member

- Feb 21, 2013

- 739

This is an old exam.

What real x satisfies equation \(\displaystyle 4\cos^2(x)-4=6\cos(x)\)

Progress:

Subsitute \(\displaystyle u=\cos(x)\) and I solve this equation

\(\displaystyle 4u^2-6u-4=0 \)

\(\displaystyle u_1=-\frac{1}{2}\) and \(\displaystyle u_2=2\)

so if we take arccos of them we get

\(\displaystyle x=\frac{3\pi}{2}+2k\pi\) which agree with facit but they got also \(\displaystyle x=-\frac{3\pi}{2}+2k\pi\) which I dont understand also how shall I know what \(\displaystyle x=\cos^{-1}(2)\) is in exam? I am doing something wrong or..?

Regards,

\(\displaystyle |\pi\rangle\)