# Solving trig equations

• Jan 13th 2010, 02:06 PM
Paymemoney
Solving trig equations
Hi
I need help on solving some of the following equations:
1) Solve $cos^2(x)-cos(x)sin(x)$ for x = $[0,2\pi]$.

2)Solve sec(x)=2.5 in the interval $[-\pi,\pi]$, giving the answers correct to two decimal places.

P.S
• Jan 13th 2010, 02:19 PM
e^(i*pi)
Quote:

Originally Posted by Paymemoney
Hi
I need help on solving some of the following equations:
1) Solve $cos^2(x)-cos(x)sin(x)$ for x = $[0,2\pi]$.

2)Solve sec(x)=2.5 in the interval $[-\pi,\pi]$, giving the answers correct to two decimal places.

P.S

1. What is this expression equal to? If it's equal to 0 then factor

$cos(x)(cos(x)-sin(x)) = 0$

$cos(x)=0$ or $cos(x) = sin(x)$

2.Take the reciprocal of both sides

$cos(x) = \frac{2}{5}$
• Jan 13th 2010, 02:23 PM
Paymemoney
yeh the first question is equal to 0
• Jan 13th 2010, 02:43 PM
Paymemoney
in question 2 I'm not sure if i have done it correct:
$sec(x)=\frac{1}{cos(x)}$
so $\frac{1}{\frac{2}{5}}=2.5*\frac{2}{5}$
• Jan 14th 2010, 02:33 AM
Paymemoney
Quote:

Originally Posted by e^(i*pi)
1. What is this expression equal to? If it's equal to 0 then factor

$cos(x)(cos(x)-sin(x)) = 0$

$cos(x)=0$ or $cos(x) = sin(x)$

i get two solutions for when cos(x)=0, which are $\frac{\pi}{2}$ & $\frac{3\pi}{2}$.
However how do you get the solutions for cos(x)=sin(x)
• Jan 14th 2010, 02:38 AM
Prove It
Quote:

Originally Posted by Paymemoney
i get two solutions for when cos(x)=0, which are $\frac{\pi}{2}$ & $\frac{3\pi}{2}$.
However how do you get the solutions for cos(x)=sin(x)

Divide both sides by $\cos{x}$.

So $\tan{x} = 1$.