Quick help - Trigonometric equations:

How do you solve trigonometic equations such as secØ=3 where 0<Ø<2pi

I know you use the trigonometric functions where you do the inverse of cos(1/3)
But how do you find out the other value without drawing a graph?
How do you know if you have to take it away from 2pi or just pi, or add it to pi to get the other value? :\

Reply 1

nuodai

17

Well

\sec \theta = 3

if and only if

\cos \theta = \dfrac{1}{3}

, so the set of values of

\theta

in the range

0 \le \theta \le 2\pi

which satisfy the equation is the same in both cases. So you don't need to worry about adding, subtracting, etc... you can just solve the cosine equation.

For what it's worth, though, the same rules apply for sec as for cos. For example

\cos (-\theta) = \cos \theta \implies \dfrac{1}{\cos (-\theta)} = \dfrac{1}{\cos \theta}

, and so

\sec (-\theta) = \sec \theta

. Similarly,

\sec (\theta + \pi) = -\sec \theta

... and so on.

(edited 13 years ago)

Reply 2

dnumberwang

12

You can always sketch out a little graph in the margin or somewhere if it helps you, I do that sometimes

Reply 3

Nkhan

OP

Original post by nuodai

Well

\sec \theta = 3

if and only if

\cos \theta = \dfrac{1}{3}

, so the set of values of

\theta

in the range

0 \le \theta \le 2\pi

which satisfy the equation is the same in both cases. So you don't need to worry about adding, subtracting, etc... you can just solve the cosine equation.

For what it's worth, though, the same rules apply for sec as for cos. For example