Trigonometry - How to solve this equation and find the solutions.

[mirandab17]

Solve: sin^2x = (sinx)(cosx)
0 < x < 2pi

Find exact solutions.

Okay, so I got 0, pi/4, but just wasn't sure how to do this...

I know you bring the sinxcosx to the other side, making it

sin^2x - sinxcosx = 0

...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on Earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!

 

[Curious3141]

Solve: sin^2x = (sinx)(cosx)
0 < x < 2pi

Find exact solutions.

Okay, so I got 0, pi/4, but just wasn't sure how to do this...

I know you bring the sinxcosx to the other side, making it

sin^2x - sinxcosx = 0

...and then I tried factoring out sinx leaving me with sinx(sinx - cosx). That's where I got lost. So sinx is 0 at 0 and pi on the unit circle, so two solutions are 0 and pi, but what on Earth do I do with sinx - cosx? That's when I started thinking about possible other identities that could help me out, but each left me with more craziness on my page!

\sin x(\sin x - \cos x) = 0

If a*b=0 then either a = 0 OR b = 0 (or both can be zero).

You've covered the first part (equating \sin x to zero) and found all the solutions).

For the second part, equate \sin x - \cos x = 0.

The easiest way to go from here is to divide throughout by \cos x then rearrange the equation. But take care that by doing this, you're disallowing solutions that make \cos x = 0 (because then you'd be dividing by 0). So check that none of those satisfy the original equation.

EDIT: In this case, it's not strictly necessary to do this (the check I mentioned), but it's good practice when an equation is solved by cancelling out factors from both sides (rather than factorising the entire expression).

 

[SammyS]

Another way to solve

  cos(x) - sin(x) = 0

is to multiply both sides by cos(x) + sin(x). Then recognize the identity for cos(2x). With this method you have to check for extraneous solutions.

(Next someone will suggest the exact way to solve this.)

 

[Curious3141]

(Next someone will suggest the exact way to solve this.)

Well, OK, if you twist my arm.

\sin x - \cos x = 0 \Rightarrow \sqrt{2}\sin(x - \frac{\pi}{4}) = 0 \Rightarrow \sin(x - \frac{\pi}{4}) = 0

 

[ehild]

The equation sinx-cosx =0 is equivalent with sinx=cosx. There are two angles between 0 and 2pi having equal sine and cosine.

ehild