Re: Trigonometric Functions

It is odd that the stated interval for $\displaystyle x$ is given in radians, yet the roots are given in degrees. Anyway, we are told:

$\displaystyle 0\le x\le 180^{\circ}$ which means:

$\displaystyle 0\le 2x\le 360^{\circ}$

From the given equation, we find:

$\displaystyle 2x=45^{\circ},\,135^{\circ}$ hence:

$\displaystyle x=22.5^{\circ},\,67.5^{\circ}$

Re: Trigonometric Functions

Ok, thank you. My only other question then is why, based on the 2 answer outcome to the problem in the first post, is the answer to the similar question of $\displaystyle \cos3x=-\frac{1}{2}$ for $\displaystyle 0\leq x\leq180$ shown as a three solution answer$\displaystyle x=40,80,160$?

Following, on from above I get:

$\displaystyle \cos3x=\frac{1}{2}\rightarrow3x=60$. Cos is negative in the 2nd & 3rd quadrants.

$\displaystyle 3x=180-60=120$ or $\displaystyle 3x=180+60=240$

$\displaystyle x=40,80$

How do I account for the 160?

Re: Trigonometric Functions

With:

$\displaystyle 0\le x\le 180^{\circ}$

we have:

$\displaystyle 0\le 3x\le 540^{\circ}$

So, we find:

$\displaystyle 3x=120^{\circ},\,240^{\circ},\,480^{\circ}$ and so:

$\displaystyle x=40^{\circ},\,80^{\circ},\,160^{\circ}$

Re: Trigonometric Functions