# Trigonometric Functions

• Nov 12th 2012, 09:03 AM
GrigOrig99
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 $\displaystyle 180^{\circ}$ is? I have a feeling my approach is slightly incorrect here. Can anyone help me clear this up?

Many thanks.

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

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

Ans.
(From text book): $\displaystyle 22.5^{\circ},67.5^{\circ}$
• Nov 12th 2012, 10:30 AM
MarkFL
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:

$\displaystyle x=22.5^{\circ},\,67.5^{\circ}$
• Nov 12th 2012, 10:50 AM
GrigOrig99
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?
• Nov 12th 2012, 11:03 AM
MarkFL
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:

$\displaystyle x=40^{\circ},\,80^{\circ},\,160^{\circ}$
• Nov 12th 2012, 11:21 AM
GrigOrig99
Re: Trigonometric Functions
Ok, thank you.