1. ## solving 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!

2. Originally Posted by jvignacio
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

$\displaystyle sin(x) = 0 ~when ~ x = 0 ~or ~x= \pi ~or~x= -\pi$
$\displaystyle sin(x) = 1/2 ~when ~x = \pi/6~ or ~x = 5\pi/6$

3. Originally Posted by jvignacio
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!
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 +-$\displaystyle \pi$

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

$\displaystyle sin(x) = 0 ~when ~ x = 0 ~or ~x= \pi ~or~x= -\pi$
$\displaystyle sin(x) = 1/2 ~when ~x = \pi/6~ or ~x = 5\pi/6$
Hi, how did you simplify $\displaystyle cos^2x+cosx$ ?

5. Originally Posted by jvignacio
Hi, how did you simplify $\displaystyle cos^2x+cosx$ ?

$\displaystyle cos^2(x) + cos(x) = cos(x)\times cos(x) + cos(x)$

Take cos(x) as common thus

cos(x)*(cos(x)+1) = 0

Now either of the two terms is zero

so cos(x) = 0 or cos(x) = -1

$\displaystyle cos^2(x) + cos(x) = cos(x)\times cos(x) + cos(x)$

Take cos(x) as common thus

cos(x)*(cos(x)+1) = 0

Now either of the two terms is zero

so cos(x) = 0 or cos(x) = -1
okay so you have $\displaystyle cos(x) = 0$ and $\displaystyle cos(x) = -1$

now what do u do with the $\displaystyle -\pi <= x <= \pi$ ?

7. Originally Posted by jvignacio
okay so you have $\displaystyle cos(x) = 0$ and $\displaystyle cos(x) = -1$

now what do u do with the $\displaystyle -\pi <= x <= \pi$ ?
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