1. ## Trigonometric Functions

The final line of my attempt contains the 2 answers listed in the book, but I'm not sure what the basis for excluding all the other values under $180^{\circ}$ is? I have a feeling my approach is slightly incorrect here. Can anyone help me clear this up?

Many thanks.

Q.
Solve the equation $\sin2x=\frac{1}{\sqrt{2}}$ for $0\leq x\leq\pi$.

Attempt: $\sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=45^{\circ}$. Sin is positive in the 1st & 2nd quadrants.
$\sin2x=\frac{1}{\sqrt{2}}\rightarrow2x=0^{\circ}+4 5^{\circ}=45^{\circ}$ or $180^{\circ}-45^{\circ}=135^{\circ}$
Since the angle is $2x$ add $90^{\circ}$ & $270^{\circ}$ to each angle (i.e. $2(45^{\circ})$ & $2(135^{\circ}$):
$2x=45^{\circ},135^{\circ},315^{\circ}$ or $2x=135^{\circ},225^{\circ},405^{\circ}$
$x=22.5^{\circ},67.5^{\circ},157.5^{\circ}$ or $x=67.5^{\circ},112.5^{\circ},202.5^{\circ}$

Ans.
(From text book): $22.5^{\circ},67.5^{\circ}$

2. ## Re: Trigonometric Functions

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

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

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

From the given equation, we find:

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

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

3. ## 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 $\cos3x=-\frac{1}{2}$ for $0\leq x\leq180$ shown as a three solution answer $x=40,80,160$?

Following, on from above I get:
$\cos3x=\frac{1}{2}\rightarrow3x=60$. Cos is negative in the 2nd & 3rd quadrants.
$3x=180-60=120$ or $3x=180+60=240$
$x=40,80$
How do I account for the 160?

4. ## Re: Trigonometric Functions

With:

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

we have:

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

So, we find:

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

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

5. ## Re: Trigonometric Functions

Ok, thank you.