• Apr 15th 2013, 06:33 PM
choi105
Determine all the possible values where 0 degree <= x <= 360 degree such that;

sin x sec x - 3 sin x = 0
• Apr 15th 2013, 06:40 PM
chiro
Hey choi105.

Please show us what you have tried.
• Apr 15th 2013, 06:55 PM
choi105
sin x sec x - 3 sin x = 0
sin x 1/cos x - 3 sin x = 0
sin x - 3 sin x = cos x

then i not sure the step correct or not...
• Apr 15th 2013, 07:45 PM
chiro
Hint: Try thinking in terms of tan(x).
• Apr 15th 2013, 08:04 PM
Prove It
Quote:

Originally Posted by choi105
sin x sec x - 3 sin x = 0
sin x 1/cos x - 3 sin x = 0
sin x - 3 sin x = cos x

then i not sure the step correct or not...

First of all, your third line is incorrect, since when you multiply everything through by $\cos{(x)}$ you're supposed to get $\displaystyle \sin{(x)} - 3\sin{(x)}\cos{(x)} \ 0$.

Anyway, you should factorise the entire expression first and set each factor equal to 0.

\displaystyle \begin{align*} \sin{(x)}\sec{(x)} - 3\sin{(x)} &= 0 \\ \sin{(x)} \left[ \sec{(x)} - 3 \right] &= 0 \\ \sin{(x)} = 0 \textrm{ or } \sec{(x)} - 3 &= 0 \end{align*}

Go from here...