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!

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- Mar 16th 2009, 05:44 AMjvignaciosolving equations help
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! - Mar 16th 2009, 05:56 AMADARSH
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$ - Mar 16th 2009, 05:57 AMe^(i*pi)
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 +-$\displaystyle \pi$ - Mar 17th 2009, 12:52 AMjvignacio
- Mar 17th 2009, 01:05 AMADARSH
- Mar 17th 2009, 04:56 AMjvignacio
- Mar 17th 2009, 05:09 AMADARSH
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