Two geometry equations

**totalnewbie** · March 20th 2006, 01:04 AM

How ?

**ThePerfectHacker** · March 20th 2006, 02:43 PM

For Problem 1:
Factor to get, call $y=x-\frac{\pi}{5}$
$\cos y (2\cos^2y-3\cos y-1)=0$
Thus,
$\cos y=0$ or $2\cos^2y-3\cos y-1=0$
Solving the second equation with quadradic formula:
$\cos y=\frac{3\pm\sqrt{17}}{4}$
Thus,
$\cos y=0$
$\cos y\approx 1.78$ --Impossible.
$\cos y\approx -.28$
Solve for $y$
after that you can solve for $x$

**ThePerfectHacker** · March 20th 2006, 02:49 PM

The product formula,
$\sin x\sin y=\frac{1}{2}[\cos(x-y)-\cos(x+y)]$
Thus,
$\sin (\pi/4+x)\sin (\pi/4-x)$
Can be expressed as,
$\frac{1}{2}[\cos(2x)-\cos(\pi/2)]$
But $\cos(\pi/2)=0$ thus,
$\frac{1}{2}\cos 2x=.33$
Thus,
$\cos 2x=.66$
Now you can solve for $2x$ for which you can solve for $x$

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