Conditional Trigonometric Equations

**senna** · February 7th 2008, 02:41 PM

$4sin^2x-3=0$

How do I find the zeros in exact values?

**Soroban** · February 7th 2008, 03:24 PM

Hello, senna!

You've never seen one of these before?

Solve for $x\!:\;\;4\sin^2\!x-3\:=\:0$

Just use your knowledge of algebra . . . then Trig at the end.

Add 3 to both sides: . $4\sin^2\!x \:=\:3$

Divide by 4: . $\sin^2\!x \:=\:\frac{3}{4}$

Take square roots: . $\sin x \:=\:\pm\frac{\sqrt{3}}{2}$

Can you finish it now?

**senna** · February 12th 2008, 07:08 PM

Thanks for the help Soroban. It's practically instant now.

$ \frac{{\pi}}{3} $ , $ \frac{{2\pi}}{3} $ , $ \frac{{4\pi}}{3} $ , $ \frac{{5\pi}}{3} $

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