# Solving Trigonometric Equation

• May 16th 2011, 04:09 PM
trigzyo
Solving Trigonometric Equation
Hey, new here and need some help with a homework question.

tan(x) + sec(2x) = 1

Ive tried 3 different substitutions/methods but I can't get to a solution. Thanks fo rthe help :D
• May 16th 2011, 04:45 PM
Sudharaka
Quote:

Originally Posted by trigzyo
Hey, new here and need some help with a homework question.

tan(x) + sec(2x) = 1

Ive tried 3 different substitutions/methods but I can't get to a solution. Thanks fo rthe help :D

Dear trigzyo,

Obviously $x=0$ and $x=\pi$ are solutions to this equation. More generally any constant multiple of $\pi$ is a solution.

That is, $x=n\pi \mbox{~where~} n\in Z$
• May 16th 2011, 04:48 PM
trigzyo
thanks, I already knew the actual solutions because they are in the back of the book. I was looking for an algebraic solution to find the solutions. Thanks a bunch though!
• May 16th 2011, 08:37 PM
Soroban
Hello, trigzyo!

Welcome aboard!

I solved it . . . but it took me a while.
Hope you can following my reasoning.

Quote:

$\tan x + \sec2x \:=\: 1$

I will assume: . $0 \,\le\,x\,<\,2\pi$

We have: . $\frac{\sin x}{\cos x} + \frac{1}{\cos2x} \;=\;1$

Multiply the first fraction by $\frac{\cos x}{\cos x}\!:\;\;\frac{\sin x\cos x}{\cos^2\!x} + \frac{1}{\cos2x} \;=\;1$

Then we have: . $\frac{\frac{1}{2}\sin2x}{\frac{1+\cos2x}{2}} + \frac{1}{\cos2x} \;=\;1 \quad\Rightarrow\quad \frac{\sin2x}{1+\cos2x} + \frac{1}{\cos2x} \;=\;1$

Multiply by $\cos2x(1+\cos2x)\!:\;\;\sin2x\cos2x + 1 + \cos2x \;=\;\cos2x(1+\cos2x)$

. . $\sin2x\cos2x + 1 \;=\;\cos^2\!x \quad\Rightarrow\quad \sin2x\cos2x + 1 \;=\;1 -\sin^2\!x$

. . $\sin^2\!2x + \sin2x\cos2x \;=\;0 \quad\Rightarrow\quad \sin2x(\sin2x + \cos2x) \;=\;0$

We have two equations to solve . . .

$\sin2x \:=\:0 \quad\Rightarrow\quad 2x \:=\:0,\:\pi,\:2\pi,\:3\pi \quad\Rightarrow\quad \boxed{x \:=\:0,\:\frac{\pi}{2},\:\pi,\:\frac{3\pi}{2}}$

$\sin2x + \cos2x \:=\:0 \quad\Rightarrow\quad \sin2x \:=\:-\cos2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:-1$
. . $\tan2x \:=\:-1 \quad\Rightarrow\quad 2x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi }{4},\:\frac{15\pi}{4}$
. . $\boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi }{8},\:\frac{15\pi}{8}}$

• May 17th 2011, 01:20 AM
Sudharaka
Quote:

Originally Posted by Soroban
Hello, trigzyo!

Welcome aboard!

I solved it . . . but it took me a while.
Hope you can following my reasoning.

I will assume: . $0 \,\le\,x\,<\,2\pi$

We have: . $\frac{\sin x}{\cos x} + \frac{1}{\cos2x} \;=\;1$

Multiply the first fraction by $\frac{\cos x}{\cos x}\!:\;\;\frac{\sin x\cos x}{\cos^2\!x} + \frac{1}{\cos2x} \;=\;1$

Then we have: . $\frac{\frac{1}{2}\sin2x}{\frac{1+\cos2x}{2}} + \frac{1}{\cos2x} \;=\;1 \quad\Rightarrow\quad \frac{\sin2x}{1+\cos2x} + \frac{1}{\cos2x} \;=\;1$

Multiply by $\cos2x(1+\cos2x)\!:\;\;\sin2x\cos2x + 1 + \cos2x \;=\;\cos2x(1+\cos2x)$

. . $\sin2x\cos2x + 1 \;=\;\cos^2\!x \quad\Rightarrow\quad \sin2x\cos2x + 1 \;=\;1 -\sin^2\!x$

. . $\sin^2\!2x + \sin2x\cos2x \;=\;0 \quad\Rightarrow\quad \sin2x(\sin2x + \cos2x) \;=\;0$

We have two equations to solve . . .

$\sin2x \:=\:0 \quad\Rightarrow\quad 2x \:=\:0,\:\pi,\:2\pi,\:3\pi \quad\Rightarrow\quad \boxed{x \:=\:0,\:\frac{\pi}{2},\:\pi,\:\frac{3\pi}{2}}$

$\sin2x + \cos2x \:=\:0 \quad\Rightarrow\quad \sin2x \:=\:-\cos2x \quad\Rightarrow\quad \frac{\sin2x}{\cos2x} \:=\:-1$
. . $\tan2x \:=\:-1 \quad\Rightarrow\quad 2x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi }{4},\:\frac{15\pi}{4}$
. . $\boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi }{8},\:\frac{15\pi}{8}}$

Dear Soroban,

Quote:

$\frac{\sin x}{\cos x} + \frac{1}{\cos2x} \;=\;1;~{\cos x}\neq 0\Rightarrow x\neq{2k\pi\pm \frac{\pi}{2}}\mbox{~where~}k\in Z$
Hence your first set of solutions should be,

$x=.........-2\pi,-\pi,0,\pi,2\pi.......$