$\displaystyle 4sin^2x-3=0$

How do I find the zeros in exact values?

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- Feb 7th 2008, 02:41 PMsennaConditional Trigonometric Equations
$\displaystyle 4sin^2x-3=0$

How do I find the zeros in exact values? - Feb 7th 2008, 03:24 PMSoroban
Hello, senna!

You've never seen one of these before?

Quote:

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

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

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

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

Can you finish it now?

- Feb 12th 2008, 07:08 PMsenna
Thanks for the help Soroban. It's practically instant now.

$\displaystyle

\frac{{\pi}}{3}

$, $\displaystyle

\frac{{2\pi}}{3}

$, $\displaystyle

\frac{{4\pi}}{3}

$, $\displaystyle

\frac{{5\pi}}{3}

$