#### Mentallic

Homework Helper

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as follows:

[tex]2sinxcosx=2sin^2x[/tex]

[tex]2sinx(sinx-cosx)=0[/tex]

Therefore,

[tex]2sinx=0[/tex] (1)

[tex]sinx-cosx=0[/tex] (2)

(2) -- [tex]cosxtanx-cosx=0[/tex]

[tex]cosx(tanx-1)=0[/tex]

Therefore,

[tex]cosx=0[/tex] (3)

[tex]tanx-1=0[/tex] (4)

Hence my solutions should be solving equations (1), (3) and (4).

i.e. [tex]sinx=0[/tex], [tex]x=0,\pi[/tex]

[tex]cosx=0[/tex], [tex]x=\pi/2[/tex]

[tex]tanx=1[/tex], [tex]x=\pi/4[/tex]

Hence, my solution set is [tex]x=0,\pi/4,\pi/2,\pi[/tex]

But testing the solutions (which I wouldn't have done so in test conditions) show that [tex]\pi/2[/tex] does not satisfy the original equation.

This means [tex]cosx\neq 0[/tex], but why?

I can't figure out where this extraneous solution came from, and which step I made was invalid to cause this. e.g. I never squared or multiplied the equation by anything etc.