# Solving Trigonometric Equations

• Nov 20th 2012, 03:16 PM
vanessa81929
Solving Trigonometric Equations
Solve the following equation algebraically. Give a general solution.

3cos2x+cosx=2
• Nov 20th 2012, 03:20 PM
Plato
Re: Solving Trigonometric Equations
Quote:

Originally Posted by vanessa81929
Solve the following equation algebraically. Give a general solution.
3cos2x+cosx=2

Can you solve $\displaystyle 3u^2+u-2=0~?$
• Nov 20th 2012, 03:21 PM
skeeter
Re: Solving Trigonometric Equations
Quote:

Originally Posted by vanessa81929
Solve the following equation algebraically. Give a general solution.

3cos2x+cosx=2

$\displaystyle 3\cos^2{x} + \cos{x} - 2 = 0$

$\displaystyle (3\cos{x} - 2)(\cos{x} + 1) = 0$

can you finish?
• Nov 20th 2012, 03:26 PM
vanessa81929
Re: Solving Trigonometric Equations
cosx= 2/3 cosx= -1

what do you do next?
• Nov 20th 2012, 03:34 PM
skeeter
Re: Solving Trigonometric Equations
Quote:

Originally Posted by vanessa81929
cosx= -2/3 cosx= -1

what do you do next?

in the interval $\displaystyle [0, 2\pi]$ ...

$\displaystyle \cos{x} = {\color{red}+}\frac{2}{3}$ ... $\displaystyle x = \arccos\left(\frac{2}{3}\right)$ and $\displaystyle x = 2\pi - \arccos\left(\frac{2}{3}\right)$

$\displaystyle \cos{x} = -1$ ... you tell me what x value on the unit circle has a cosine of -1

if you want all solutions, add integer multiples of $\displaystyle 2\pi$ to all three solutions.
• Nov 20th 2012, 03:43 PM
vanessa81929
Re: Solving Trigonometric Equations
For the first one, x= 0.841+2npi

And for the second one, x=pi+2npi

Where are you getting a third solution from?
• Nov 20th 2012, 03:54 PM
skeeter
Re: Solving Trigonometric Equations
$\displaystyle \cos{x} = \frac{2}{3}$ has two solutions, one in quad I and one in quad IV

for $\displaystyle x = \pi$, the general solution should be written ...

$\displaystyle x = \pi + 2k\pi , k \in \mathbb{Z}$

... and put away the calculator.