solution for trignometric equations

• Feb 2nd 2014, 10:16 PM
arangu1508
solution for trignometric equations
Dear friends,

The following is the question, I encountered:

What is the number of solutions for the pair of following equations

2 sin2x - cos2x = 0
2 cos2x - 3sinx = 0

in the interval [0,2Pi]

I added both the equations and I got 2sin2x - 3sinx +1 = 0

For which I thought the solution was 300, 900 and 1500 as the answers. But they say it is wrong?!

Is it so? where I went wrong and what is the correct answer?

could anyone help me.

with warm regards,

Aranga

• Feb 3rd 2014, 12:00 AM
Prove It
Re: solution for trignometric equations
If your interval is \displaystyle \displaystyle \begin{align*} \left[ 0, 2\pi \right] \end{align*}, meaning your angle is measured in radians, why have you written your answer in degrees?
• Feb 3rd 2014, 12:36 AM
arangu1508
Re: solution for trignometric equations
Sir,

Then it is Pi/2, Pi/6 and 5Pi/6. Is it ok?

Whether my answer is correct now?

with warm regards,

Aranga
• Feb 3rd 2014, 12:39 AM
Prove It
Re: solution for trignometric equations
Yes that's fine now :)
• Feb 3rd 2014, 12:50 AM
BobP
Re: solution for trignometric equations
? $\displaystyle \pi/2$ doesn't satisfy either equation.
• Feb 3rd 2014, 12:55 AM
arangu1508
Re: solution for trignometric equations
when we solve the equation, pi/2 is one of the solutions. But I don't where I went wrong? Seems to be straight forward but I do not able to locate my mistake.
• Feb 3rd 2014, 06:17 AM
BobP
Re: solution for trignometric equations
You can't add the two equations together and expect to retain the solutions to the original equations.

They are only simultaneous in the sense that you are looking for solutions common to both equations, (or at least that's what I'm assuming is required). In that case, solve the two separately.

To illustrate the first point, consider the equations $\displaystyle 1+3x=0$ and $\displaystyle 2-6x=0.$
They have, respectively, solutions $\displaystyle x=-1/3$ and $\displaystyle x=1/3.$
Add them and you get$\displaystyle 3-3x=0$ which has the solution $\displaystyle x=1.$
Both 'original' solutions have been lost and a spurious one created.