# Math Help - Trig Help Please!!!

I don't know what to do. Can someone please help me solve these. I don't even know how to start.

1. Find all solutions in the interval [0, 2pie) of 2cos(x)-square root 3=0

2. Find all solutions in the interval [0, 2pie) of sin(x)=(1/4sin(x))

3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)

4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0

2. Originally Posted by jrose011
I don't know what to do. Can someone please help me solve these. I don't even know how to start.

1. Find all solutions in the interval [0, 2pie) of 2cos(x)-square root 3=0

2. Find all solutions in the interval [0, 2pie) of sin(x)=(1/4sin(x))

3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)

4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0

(1) 2cosx-\sqrt{3}=0

cosx=\frac{\sqrt{3}{2}}

x=30 , 330

(2) $sinx = \frac{1}{4sinx }$

$
4sin^2x=1
$

sinx =1/2 or sinx =-1/2

3. Hello jrose011
Originally Posted by jrose011
...

3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)
With an equation like

$3\sec^2x\tan x = 4\tan x$

you have to be careful that you don't miss any solutions by dividing both sides by $\tan x$, to get:

$3\sec^2x = 4$

Remember the first commandment of arithmetic, which is "Thou shalt not divide by zero."

You see, $\tan x = 0$ is a possible solution, isn't it? (Because $3\sec^2x \times 0 = 0 = 4 \times 0$.) And if you divide both sides by it, you'll lose this solution.

So you have to say:

$3\sec^2x\tan x = 4\tan x$

$\Rightarrow \tan x = 0$ or $3\sec^2 x = 4$

So you now have 2 equations to solve. $\tan x = 0$ is very easy. Can you do the second one? Write it as:

$\frac{3}{\cos^2x} = 4$

$\Rightarrow \cos^2x = \frac{3}{4}$

... and then don't forget the second commandment of arithmetic "When thou takest square roots of both sides, remember thy plus-or-minus sign".

4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0
Use $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\sin(A-B) = \sin A \cos B - \cos A \sin B$:

$\sin(x + \tfrac{\pi}{4}) + \sin(x - \tfrac{\pi}{4}) = 0$

$\Rightarrow \sin x \cos\tfrac{\pi}{4} + \cos x \sin\tfrac{\pi}{4} +\sin x \cos\tfrac{\pi}{4} - \cos x \sin\tfrac{\pi}{4} =0$

$\Rightarrow 2\sin x \cos\tfrac{\pi}{4} = 0$

$\Rightarrow \sin x = 0$

$\Rightarrow x = \dots$