# Solving Trigonometric equations

• Mar 23rd 2009, 10:36 AM
mj.alawami
Solving Trigonometric equations
Question

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

Attempt

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

$\displaystyle t=2 ;t=-1$

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

Is this correct?

Thank you
• Mar 23rd 2009, 10:48 AM
stapel
Quote:

Originally Posted by mj.alawami
$\displaystyle 2sin^2t-cost-1=0$ if $\displaystyle 0 \leq \pi \leq 2\pi$

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:

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

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

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

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

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

...and so forth. (Wink)