# Thread: Solving Trigonometric Conditional Equations

1. ## 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?

2. ## Re: Solving Trigonometric Conditional Equations

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 \begin{align*} \sin{x} = 0 \end{align*}.

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

\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)...

3. ## 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.

4. ## Re: Solving Trigonometric Conditional Equations

If we are ruling out pi/2 and 3pi/2 that of course just leaves 0 and pi

5. ## Re: Solving Trigonometric Conditional Equations

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 \begin{align*} x = \frac{\pi}{4} \end{align*} be a solution? It doesn't make \displaystyle \begin{align*} \cos{x} = 0 \end{align*}...

6. ## Re: Solving Trigonometric Conditional Equations

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

7. ## Re: Solving Trigonometric Conditional Equations

Many apologies, I misread the question! I've been solving tan2x=sin2x

8. ## Re: Solving Trigonometric Conditional Equations

Originally Posted by biffboy
The original equation was tan2x=sin2x and x=pi/4 is not a solution.
No, it was \displaystyle \begin{align*} \tan^2{x} = \sin{2x} \end{align*}.

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