given that $\displaystyle -\pi <= x <= \pi$
solve:
a) $\displaystyle cos^2x+cosx=0$
b) $\displaystyle 2sin^2x - sinx = 0$
NOTE: <= (less than or equal to)
any help much appreciated!
cos(x)(1+cos(x)) = 0
Either $\displaystyle cos(x) = 0$ or $\displaystyle 1+cos(x) = 0 $
$\displaystyle
cos(x) = 0~ when~ x = \pi/2 ~or~ - \pi/2 $
$\displaystyle cos(x) = -1~ when ~x= \pi~ or~ x = - \pi $
2sin^2(x) - sin(x) = 0
sin(x) (2sin(x)-1) = 0
either sin(x) = 0 or sin(x) = 1/2
$\displaystyle sin(x) = 0 ~when ~ x = 0 ~or ~x= \pi ~or~x= -\pi$
$\displaystyle
sin(x) = 1/2 ~when ~x = \pi/6~ or ~x = 5\pi/6$
You can factorise both of these:
$\displaystyle cos(x)(cos(x) + 1) = 0$ so $\displaystyle cos(x) = 0$ or $\displaystyle cos(x) = -1$
and the same with sin(x). Just remember the points at -1 will repeat every +-[tex]2\pi/MATH] and the points at 0 will repeat every +-$\displaystyle \pi$
When is cos(x) = 0 or when is cos(x) = -1
What are the values of x when this values are satisfied
For x= 0 This happens when x= 90degree or x = - 90 degrees
And for cos(x)= -1 this happens when x= 180degree or x= -180 degree
This is done using the property that
cos(-x) =cos(x)
ie; if the answer is 90 degree than answer is also -90 degree as this value also lies in the domain
Read this
http://en.wikipedia.org/wiki/Trigonometric_function