# Solving Trigonometric Conditional Equations

• Apr 16th 2012, 10:44 AM
ostheimerdl
Solving Trigonometric Conditional Equations
tan^2x=sin2x, 0<=x<2pi

cosx cannot = 0 ; x cannot = pi/2 or 3pi/2

sin^2x/cos^2x=2sinxcosx
sin^2x=2sinxcos^3x
sin^2x-2sinxcos^3x=0
sinx(sinx-2cos^3x)=0
sinx=0 or sinx-2cos^3x=0
+-sqrt(1-cos^2x) - 2cos^3x=0
+-sqrt(1-cos^2x) = 2cos^3x
*square both sides to remove radical
1-cos^2x = 4cos^6x
4cos^6x-cos^2x-1=0

I don't know how to solve this equation, can someone please provide me with a link to read how to solve this equiation?
• Apr 16th 2012, 06:45 PM
Prove It
Re: Solving Trigonometric Conditional Equations
Quote:

Originally Posted by ostheimerdl
tan^2x=sin2x, 0<=x<2pi

cosx cannot = 0 ; x cannot = pi/2 or 3pi/2

sin^2x/cos^2x=2sinxcosx
sin^2x=2sinxcos^3x
sin^2x-2sinxcos^3x=0
sinx(sinx-2cos^3x)=0
sinx=0 or sinx-2cos^3x=0
+-sqrt(1-cos^2x) - 2cos^3x=0
+-sqrt(1-cos^2x) = 2cos^3x
*square both sides to remove radical
1-cos^2x = 4cos^6x
4cos^6x-cos^2x-1=0

I don't know how to solve this equation, can someone please provide me with a link to read how to solve this equiation?

First of all, don't forget that you need to solve \displaystyle \displaystyle \begin{align*} \sin{x} = 0 \end{align*}.

Anyway, once you get to \displaystyle \displaystyle \begin{align*} \sin{x} - 2\cos^3{x} = 0 \end{align*} I would do this...

\displaystyle \displaystyle \begin{align*} \sin{x} - 2\cos^3{x} &= 0 \\ \sin{x} &= 2\cos^3{x} \\ \frac{\sin{x}}{\cos{x}} &= 2\cos^2{x} \textrm{ we can do this since it has already been stated that }\cos{x} \neq 0 \\ \tan{x} &= \frac{2}{\sec^2{x}} \\ \tan{x} &= \frac{2}{\tan^2{x} + 1} \\ \tan{x}\left(\tan^2{x} + 1\right) &= 2 \\ \tan^3{x} + \tan{x} &= 2 \\ \tan^3{x} + \tan{x} - 2 &= 0 \\ X^3 + X - 2 &= 0 \textrm{ if we let }X = \tan{x} \\ \left(X - 1\right)\left(X^2 + X + 2\right) &= 0 \end{align*}

You should be able to solve this now. Also please advise if you only want real solutions or if you want complex solutions as well (they are, as the name suggests, more complex)...
• Apr 16th 2012, 09:54 PM
biffboy
Re: Solving Trigonometric Conditional Equations
tanx=1 gives x=pi/4 which cannot be a solution
I suggest the following sin2x/cos2x=sin2x sin2x=sin2xcos2x sin2x-sin2xcos2x =0 sin2x(1-cos2x)=0 sin2x=0 or 1-cos2x=0 cos2x=1

The first gives 2x=0 or pi or 2pi or 3pi So x= 0 or pi/2 or pi or3pi/2 cos2x=1 gives 2x= 0 or 2pi so x= 0 or pi which we already have.
• Apr 16th 2012, 10:58 PM
biffboy
Re: Solving Trigonometric Conditional Equations
If we are ruling out pi/2 and 3pi/2 that of course just leaves 0 and pi
• Apr 17th 2012, 02:50 AM
Prove It
Re: Solving Trigonometric Conditional Equations
Quote:

Originally Posted by biffboy
tanx=1 gives x=pi/4 which cannot be a solution
I suggest the following sin2x/cos2x=sin2x sin2x=sin2xcos2x sin2x-sin2xcos2x =0 sin2x(1-cos2x)=0 sin2x=0 or 1-cos2x=0 cos2x=1

The first gives 2x=0 or pi or 2pi or 3pi So x= 0 or pi/2 or pi or3pi/2 cos2x=1 gives 2x= 0 or 2pi so x= 0 or pi which we already have.

Why can't \displaystyle \displaystyle \begin{align*} x = \frac{\pi}{4} \end{align*} be a solution? It doesn't make \displaystyle \displaystyle \begin{align*} \cos{x} = 0 \end{align*}...
• Apr 17th 2012, 03:12 AM
biffboy
Re: Solving Trigonometric Conditional Equations
The original equation was tan2x=sin2x and x=pi/4 is not a solution.
• Apr 17th 2012, 03:17 AM
biffboy
Re: Solving Trigonometric Conditional Equations
Many apologies, I misread the question! I've been solving tan2x=sin2x
• Apr 17th 2012, 03:21 AM
Prove It
Re: Solving Trigonometric Conditional Equations
Quote:

Originally Posted by biffboy
The original equation was tan2x=sin2x and x=pi/4 is not a solution.

No, it was \displaystyle \displaystyle \begin{align*} \tan^2{x} = \sin{2x} \end{align*}.