# Trigonometric Solutions

1. Aug 12, 2017

1. The problem statement, all variables and given/known data
The number of solutions of the equation sin2x –2cosx + 4sinx = 4 in the interval [0, 5π] is what?

2. Relevant equations
sin2x=2sinxcosx

3. The attempt at a solution
sin2x-2cosx+4sinx=4
⇒2sinxcosx-2cosx+4sinx=4
⇒sinxcosx-cosx+2sinx=2
⇒cosx (sinx-1)=2-2sinx
⇒cosx (sinx-1)=2 (1-sinx)
⇒cosx (sinx-1)= -2 (sinx-1)
⇒cosx = -2
∴ There is no solution

2. Aug 12, 2017

### Orodruin

Staff Emeritus
Your last step assumes something that is not necessarily true.

3. Aug 12, 2017

### Staff: Mentor

WolframAlpha finds at least one solution in [0,π]...

4. Aug 13, 2017

### scottdave

Do you have the ability to plot the expression on the left side (with a calculator or spreadsheet)? Then see if it crosses 4 anywhere. That is a start to see how many solutions you are dealing with.

Like @Orudruin indicated - you divided both sides by an expression. There is a time when you don't want to do that.

5. Aug 14, 2017

### ElectricRay

I found indeed several with my calculator but solving it algebraically

After the step cosx (sinx-1)=2 (1-sinx) as the OP did i got lost. Im strying to solve it as well without solving it for the OP. But why you can't devide after this step, what is wrong about that?

6. Aug 14, 2017

### Orodruin

Staff Emeritus
First of all, be aware that posting complete solutions before the OP has solved the problem is not allowed. See the forum rules.

Regarding your question: In which cases is it not ok to divide by a real number $x$?