# [SOLVED] Trig equation

• Sep 17th 2007, 12:39 PM
Spec
[SOLVED] Trig equation
$\displaystyle tan(2x) = \frac{1 + sin(x)}{cos(x)}$

I'm all out of ideas here. Anyone?
• Sep 17th 2007, 01:16 PM
TKHunny
I'd start with the Right-hand side. Multiply numerator and denominator by 1 + sin(x) and see what happens.

Like many such problems, it may not lead anywhere, but it's a place to start and something else may hit you.

For example, while I was typing this, it dawned on me that I don't happen to know any formulas for tan(2x) right off the top of my head. I might be tempted to change that to sines and cosines. Maybe not.

Rule #1 on Trig Identities -- You can't break it. Just try something.
• Sep 17th 2007, 01:45 PM
Spec
Thanks, but I've already tried everything I can think of. I need to see a solution soon or my head will explode. :eek:

As if the last one wasn't bad enough, I'm now stuck with this one as well:

$\displaystyle 2cos^3(x) + sin(x) = 2cos(x)$
• Sep 17th 2007, 03:56 PM
Soroban
Hello, Spec!

Quote:

$\displaystyle 2\cos^3\!x + \sin x \:= \:2\cos x$

We have: .$\displaystyle 2\cos x(\cos^2\!x) + \sin x \;=\;2\cos x \quad\Rightarrow\quad 2\cos x(1 - \sin^2\!x) + \sin x \;=\;2\cos x$

. . . . $\displaystyle 2\cos x - 2\sin^2\!x\cos x + \sin x \;=\;2\cos x \quad\Rightarrow\quad \sin x - 2\sin^2\!x\cos x \;=\;0$

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

And we have two equations to solve:

$\displaystyle \sin x \:=\:0\quad\Rightarrow\quad\boxed{x \:=\:\pi n}$

$\displaystyle 1 - 2\sin x\cos x\:=\:0\quad\Rightarrow\quad 1 - \sin2x \:=\:0\quad\Rightarrow\quad \sin2x \:=\:1$
. . $\displaystyle 2x \:=\:\frac{\pi}{2} + 2\pi n \quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{4} + \pi n}$

• Sep 17th 2007, 04:17 PM
Soroban
Hello, Spec!

I'll give only the principal roots.
. . You can generalize the answers . . .

Quote:

Solve for $\displaystyle x\!:\;\;\tan 2x \;= \;\frac{1 + \sin x}{\cos x}$

We have: .$\displaystyle \frac{\sin2x}{\cos2x} \;=\;\frac{1+\sin x}{\cos x} \quad\Rightarrow\quad \frac{2\sin x\cos x}{1 - 2\sin^2\!x} \;=\;\frac{1+\sin x}{\cos x}$

. . $\displaystyle 2\sin x\cos^2x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . $\displaystyle 2\sin x(1 - \sin^2\!x) \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . $\displaystyle 2\sin x - 2\sin^3\!x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . which simplifies to: .$\displaystyle 2\sin^2\!x + \sin x - 1 \;=\;0$

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

And we have two equations to solve.

. . $\displaystyle 2\sin x - 1 \:=\:0\quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}$

. . $\displaystyle \sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{2}}$

• Sep 18th 2007, 02:29 AM
Spec
Quote:

Originally Posted by Soroban
Hello, Spec!

I'll give only the principal roots.
. . You can generalize the answers . . .

We have: .$\displaystyle \frac{\sin2x}{\cos2x} \;=\;\frac{1+\sin x}{\cos x} \quad\Rightarrow\quad \frac{2\sin x\cos x}{1 - 2\sin^2\!x} \;=\;\frac{1+\sin x}{\cos x}$

. . $\displaystyle 2\sin x\cos^2x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . $\displaystyle 2\sin x(1 - \sin^2\!x) \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . $\displaystyle 2\sin x - 2\sin^3\!x \;=\;1 + \sin x - 2\sin^2\!x - 2\sin^3\!x$

. . which simplifies to: .$\displaystyle 2\sin^2\!x + \sin x - 1 \;=\;0$

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

And we have two equations to solve.

. . $\displaystyle 2\sin x - 1 \:=\:0\quad\Rightarrow\quad \sin x \:=\:\frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}$

. . $\displaystyle \sin x + 1\:=\:0\quad\Rightarrow\quad \sin x \:=\:-1\quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{2}}$

Thanks! $\displaystyle x = \frac{3\pi}{2} + 2{\pi}n$ is not a solution though, since $\displaystyle cos x \neq 0$.