NEed help solving trig equations

**ThisIsFun** · December 20th 2012, 07:18 AM

I have two trig equations that i just can seem to get. Question is " Solve the trig equations on the given intervals. State any restrictions and give exact answers when posible or round to two
decimals." First one is:

$cosx=cscxtanx$

and the interval is $-\pi\leq x\leq \pi$

second problem is:

$\frac{2tan2x}{1-tan^22x}=\frac{tanx-tan3x}{1+tanxtan3x}$

and the interval is $0\leq x\leq \frac{3\pi}{2}$

**bandgeek0295** · December 20th 2012, 07:28 AM

Have you tried anything yet?

On the first problem, can we change that csc(x) and tan(x) to something a bit more managable? Perhaps sines and cosines? From there, it should be a bit clearer. On the second, can we factor anything? I always factor whenever I can, and in doing so, a trig identity that I can work with usually becomes apparent. Finally, there is one off the bat that you can work with here.

Good luck!

**Soroban** · December 20th 2012, 02:54 PM

Hello, ThisIsFun!

Solve the trig equations on the given intervals.
State any restrictions and give exact answers when possible or round to two decimals.

$\frac{2\tan2x}{1-\tan^2\!2x}\:=\:\frac{\tan x-\tan3x}{1+\tan x\tan3x}$

and the interval is: . $0\leq x\leq \tfrac{3\pi}{2}$

The right side has the form: . $\tan(A-B) \:=\:\frac{\tan A - \tan B}{1 + \tan A\tan B}$

Hence: . $\frac{\tan x-\tan3x}{1+\tan x\tan3x} \:=\:\tan(x-3x) \:=\:\tan(-2x) \;=\;-\tan2x$

The equation becomes: . $\frac{2\tan2x}{1-\tan^2\!2x} \;=\;-\tan2x$

. . $2\tan2x \:=\:-\tan2x + \tan^3\!2x \quad\Rightarrow\quad \tan^3\!2x - 3\tan2x \:=\:0$

. . $\tan2x(\tan^2\!2x - 3) \:=\:0$

Your turn . . .

**ThisIsFun** · December 20th 2012, 07:02 PM

Thank You!!!!

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