Solving Trigonometric equations

**mj.alawami** · March 23rd 2009, 10:36 AM

Question

Find the angles too?
$2sin^2t-cost-1=0$ if $0 \leq \pi \leq 2\pi$

Attempt

$cos^2t-cost-1=0$

$t=2 ;t=-1$

$arccos(2)=$ math error ; $arccos(-1)=180$

Is this correct?

Thank you

**stapel** · March 23rd 2009, 10:48 AM

Originally Posted by mj.alawami

$2sin^2t-cost-1=0$ if $0 \leq \pi \leq 2\pi$

How did "2sin^2(t)" become "cos^2(t)"...?

Instead, try using the Pythagorean Identity in the form sin^2(t) = 1 - cos^2(t), so you get:

. . . . . $2(1\, -\, \cos^2(t))\, -\, \cos(t)\, -\, 1\, =\, 0$

. . . . . $2\, -\, 2\cos^2(t)\, -\, \cos(t)\, -\, 1\, =\, 0$

. . . . . $0\, =\, 2\cos^2(t)\, +\, \cos(t)\, -\, 1$

. . . . . $0\, =\, (2\cos(t)\, -\, 1)(\cos(t)\, +\, 1)$

. . . . . $\cos(t)\, =\, \frac{1}{2}\, \mbox{ or }\, \cos(t)\, =\, -1$

...and so forth.

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