# Thread: trig solving for x

1. ## trig solving for x

1. solve for x in $\displaystyle sinx tanx+ tanx- 2sinx+ cosx=0$ for 0 <or equalx<orequal to $\displaystyle 2\pi$ rads

2. solve for x in $\displaystyle \frac{1+sinx}{cosx}+\frac{cosx}{1+sinx}=4$ for 0<or equal x <orequal $\displaystyle 2\pi$ rads

3. find the exact value of

a) $\displaystyle sin (\frac{\pi}{4}-\frac{\pi}{4})$

b) $\displaystyle sin (\frac{13\pi}{12})$

2. Hello, william!

2. Solve for $\displaystyle x\!:\;\;\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}\:=\:4\qquad \text{for }0 \leq x \leq 2\pi$

Multipy by $\displaystyle \cos x(1 + \sin x)\!:\quad (1+\sin x)^2 + \cos^2\!x \;=\;4\cos x(1 + \sin x)$

. . $\displaystyle 1 + 2\sin x + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} \;=\;4\cos x + 4\sin x\cos x$

. . . . . . . . $\displaystyle 2 + 2\sin x \;=\;4\cos x + 4\sin x\cos x$

Divide by 2: .$\displaystyle 1 + \sin x \;=\;2\cos x + 2\sin x\cos x$

Re-arrange: .$\displaystyle 2\sin x\cos x - \sin x + 2\cos x - 1 \;=\;0$

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

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

And we have:

. . $\displaystyle \sin x +1 \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-1 \quad\Rightarrow\quad x \:=\:{\color{red}\rlap{//}}\frac{3\pi}{2}$ .
extraneous root

. . $\displaystyle 2\cos x - 1 \:=\:0 \quad\Rightarrow\quad \cos x \:=\:\frac{1}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{3},\:\frac{5\pi}{3}}$

3. Find the exact value of:

$\displaystyle a)\;\;\sin\left(\frac{\pi}{4}-\frac{\pi}{4}\right)$
$\displaystyle \sin\left(\frac{\pi}{4} - \frac{\pi}{4}\right) \;=\;\sin(0) \;=\;0$

$\displaystyle b)\;\;\sin \left(\frac{13\pi}{12}\right)$
Note that: .$\displaystyle \frac{13\pi}{12} \;=\;\frac{9\pi}{12} + \frac{4\pi}{12} \;=\;\frac{3\pi}{4} + \frac{\pi}{3}$

We have: .$\displaystyle \sin\left(\frac{13\pi}{12}\right) \;=\;\sin\left(\frac{3\pi}{4} + \frac{\pi}{3}\right)$

. . . . . . $\displaystyle = \;\;\sin\left(\frac{3\pi}{4}\right)\cos\left(\frac {\pi}{3}\right) + \cos\left(\frac{3\pi}{4}\right)\sin\left(\frac{\pi }{3}\right)$

. . . . . . $\displaystyle = \;\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2} \right) + \left(-\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\ right) \;\;=\;\;\frac{1-\sqrt{3}}{2\sqrt{2}} \;\;=\;\;\frac{\sqrt{2} - \sqrt{6}}{4}$