How Many Solutions Does the Trigonometric Equation Have in the Given Interval?

[Vavi Ask]

Homework Statement

The number of solutions of the equation sin2x –2cosx + 4sinx = 4 in the interval [0, 5π] is what?

Homework Equations

sin2x=2sinxcosx

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

 

[Orodruin]

Your last step assumes something that is not necessarily true.

 

[berkeman]

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

 

[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.

 

[ElectricRay]

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.

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. I am strying to solve it as well without solving it for the OP. But why you can't divide after this step, what is wrong about that?

 

[Orodruin]

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. I am strying to solve it as well without solving it for the OP. But why you can't divide after this step, what is wrong about that?

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 $x$?