# Math Help - Trigonometric Equations

1. ## Trigonometric Equations

Hello

I have a couple of trig equations which i am having difficulty solving. I was sick for a while before the break and now that its exam review time I don't remember much. BTW this is an assignment due tomorrow, thats why its in urgent help.

Here are the questions

$3cot^2(3x-\frac{pi}{4})-1=0$
$sin2x+sinx-2cosx-1=0$
$sinx+\sqrt{3}cosx=1$ please note the cos x is not under the root

heres how i started solving the first equation =\

$3cot^2(3x-\frac{pi}{4})=1$
$cot(3x-\frac{pi}{4}=\sqrt{\frac{1}{3}}$
$tany=\frac{1}{\sqrt{\frac{1}{3}}}$
$=\sqrt{3}$
therefore
$y = \frac{2pi}{3}$ $y = \frac{pi}{3}$ $y = \frac{4pi}{3}$ $y = \frac{5pi}{3}$
now im kinda stuck. since the question does not give a limit (ie 0 to 2pi) do we need to figure out an equation or something? =\

any help appreciated
thanks
chris

2. $3\text{Cot}^2\left(3x-\frac{\pi }{4}\right)=1$
$\text{Cot}^2\left(3x-\frac{\pi }{4}\right)= \frac{1}{3}$

(Remember that $\sqrt{x^2} = |x|$)

$\left|\text{Cot}\left(3x-\frac{\pi }{4}\right)\right| = \sqrt{\frac{1}{3}}$
$\text{Cot}\left(3x-\frac{\pi }{4}\right)=\pm \sqrt{\frac{1}{3}}$
$\text{Cot}\left(3x-\frac{\pi }{4}\right)= +\sqrt{\frac{1}{3}} \text{ } \boxed{\text{I}} \text{ } , \text{ } \text{Cot}\left(3x-\frac{\pi }{4}\right)=-\sqrt{\frac{1}{3}} \text{ } \boxed{\text{II}}$

$\boxed{I.} \text{ } \text{Cot}\left(3x-\frac{\pi }{4}\right)= \sqrt{\frac{1}{3}}$
$3x-\frac{\pi }{4}+ \text{k\pi } = \frac{\pi }{3}$
$x = \frac{7\pi }{36}+\frac{\text{k\pi }}{3}$

$\boxed{II.} \text{ } \text{Cot}\left(3x-\frac{\pi }{4}\right)=-\sqrt{\frac{1}{3}}$
$3x-\frac{\pi }{4} + \text{k\pi } = \frac{2\pi }{3}$
$x = \frac{11\pi }{36}+\frac{\text{k\pi }}{3}$

So,

$x=\{ \frac{7\pi }{36}+\frac{\text{k\pi }}{3}, \frac{11\pi }{36}+\frac{\text{k\pi }}{3} \}$

3. ok i think im understanding how you got the answers,
i changed y back to the original form but how dod you come up with the later part of $x = \frac{7\pi }{36}+\frac{\text{k\pi }}{3}$
like as in why it is pi / 3
and also, do the other answers not work? or did you omit them because you get them the same way or they are redundant?

4. Hello, Chris!

Here's some help . . .

$\sin2x + \sin x - 2\cos x - 1\;=\;0$

We have: . $2\sin x\cos x + 2\sin x - 2\cos x - 1 \;=\;0$

Factor: . $\sin x(2\cos x + 1) - (2\cos x + 1)\;=\;0$

Factor: . $(2\cos x + 1)(\sin x - 1)\;=\;0$

Then we have: . $2\cos x + 1\:=\:0\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}$

. . and: . $\sin x -1\:=\:0\quad\Rightarrow\quad \sin x \:=\:1\quad\Rightarrow\quad x \:=\:\frac{\pi}{2}$

Therefore: . $x \;=\;\left\{\dfrac{2\pi}{3},\;\dfrac{4\pi}{3},\;\d frac{\pi}{2} \right\} + 2\pi n$

$\sin x+\sqrt{3}\cos x \;=\;1$

This one requires some Olympic-level gymnastics . . . and a few facts:

. . $\sin\frac{\pi}{3} \:=\:\frac{\sqrt{3}}{2},\;\cos\frac{\pi}{3} \:=\:\frac{1}{2}$ . [1]

. . $\sin A\cos B + \sin B\cos A \:=\:\sin(A + B)$ . [2]

We have: . $\sin x + \sqrt{3}\cos x \;=\;1$

Divide by 2: . $\frac{1}{2}\sin x + \frac{\sqrt{3}}{2}\cos x \;=\;\frac{1}{2}$

From [1], we have: . $\cos\frac{\pi}{3}\sin x + \sin\frac{\pi}{3}\cos x \;=\;\frac{1}{2}$

From [2], we have: . $\sin\left(x + \frac{\pi}{3}\right) \;=\;\frac{1}{2}$

Hence: . $x + \frac{\pi}{3} \:=\:\left\{\frac{\pi}{6},\;\frac{5\pi}{6}\right\} + 2\pi n$

Therefore: . $x \;=\;\left\{-\frac{\pi}{6},\;\frac{\pi}{2}\right\} + 2\pi n$

The first problem has about a dozen basic forms.
I'll try to get it straightened out later.
.