# Thread: [SOLVED] Help with Trigonometric Equations

1. ## [SOLVED] Help with Trigonometric Equations

Hi, no matter what I try, I cannot solve these. I keep on ending up with 4 terms which I cannot factor. Well anyway here are the 2 questions. Please show me how to do them.

Solve for x on the interval 0 degree </= x </= 360 degrees.
Question A) 2sinxtanx - tanx - 2sinx + 1 = 0
Question B) cosxtanx - 1 + tanx - cosx = 0

I just need to get to part where I factor.

Thanks!

2. Originally Posted by demo1
Hi, no matter what I try, I cannot solve these. I keep on ending up with 4 terms which I cannot factor. Well anyway here are the 2 questions. Please show me how to do them.

Solve for x on the interval 0 degree </= x </= 360 degrees.
Question A) 2sinx - tanx - 2sinx + 1 = 0
Question B) cosxtanx - 1 + tanx - cosx = 0

I just need to get to part where I factor.

Thanks!
Hi

Are you sure about A ? 2sinx cancel ... It remains -tanx + 1 = 0

For B : $\displaystyle \cos x \tan x - 1 + \tan x - \cos x = 0 \iff \sin x - 1 + \frac{\sin x}{\cos x} - \cos x = 0$ $\displaystyle \iff \sin x - \cos x + \frac{\sin x - \cos x}{\cos x} = 0 \iff (\sin x - \cos x) \left(1 + \frac{1}{\cos x}\right) = 0$

I think that you can finish

3. Thanks, I made a mistake copying the question. I fixed it now!

4. Originally Posted by running-gag
Hi

Are you sure about A ? 2sinx cancel ... It remains -tanx + 1 = 0

For B : $\displaystyle \cos x \tan x - 1 + \tan x - \cos x = 0 \iff \sin x - 1 + \frac{\sin x}{\cos x} - \cos x = 0$ $\displaystyle \iff \sin x - \cos x + \frac{\sin x - \cos x}{\cos x} = 0 \iff (\sin x - \cos x) \left(1 + \frac{1}{\cos x}\right) = 0$

I think that you can finish
I think i know how to solve (1/cosx - 1) for the angle but how do I solve (sinx - cosx)?

5. Originally Posted by demo1
Thanks, I made a mistake copying the question. I fixed it now!
OK

$\displaystyle 2 \sin x \tan x - \tan x - 2 \sin x + 1 = 0 \iff (2\sin x - 1) \tan x - (2\sin x - 1) = 0$$\displaystyle \iff (2\sin x - 1) (\tan x - 1) = 0$

6. Originally Posted by demo1
I think i know how to solve (1/cosx - 1) for the angle but how do I solve (sinx - cosx)?
$\displaystyle \sin x - \cos x = \sqrt{2} \:\left(\frac{\sqrt{2}}{2}\:\sin x - \frac{\sqrt{2}}{2}\:\cos x\right)$

$\displaystyle \sin x - \cos x = \sqrt{2} \:\left(\cos\left(\frac{\pi}{4}\right)\:\sin x - \sin\left(\frac{\pi}{4}\right)\:\cos x\right)$

$\displaystyle \sin x - \cos x = \sqrt{2} \:\sin\left(x-\frac{\pi}{4}\right)$

7. Wow thanks running-gag! I think I get it now.