Factorising help

val1

New member
Joined
Oct 17, 2005
Messages
40
Hi

Please can you help me to factorise the following equation to find the general solution;


\(\displaystyle 3- 2sin^2 x - 3cosx = 0\)

I multiplied the above equation by -1
\(\displaystyle 2sin^2 x + 3cosx - 3 = 0\)

and tried to factorise using the quadratic formula but I got a truly unfeasible answer
[-3 +/- (sqrt)(33)]/ 4

which makes me think that I'm not on the right track.

Thanks
 
Use the Pythagorean Identity:

. . . . .cos<sup>2</sup>(x) + sin<sup>2</sup>(x) = 1

In this case, use it in the form:

. . . . .sin<sup>2</sup>(x) = 1 - cos<sup>2</sup>(x)

Then:

. . . . .3 - 2sin<sup>2</sup>(x) - 3cos(x) = 0
. . . . .3 - 2[1 - cos<sup>2</sup>(x)] - 3cos(x) = 0
. . . . .3 - 2 + cos<sup>2</sup>(x) - 3cos(x) = 0
. . . . .2cos<sup>2</sup>(x) - 3cos(x) + 1 = 0

Does that look a little more factorable? :wink:

Eliz.
 
Hi

Please can you help me to factorise the following equation to find the general solution;


\(\displaystyle 3- 2sin^2 x - 3cosx = 0\)

I multiplied the above equation by -1
\(\displaystyle 2sin^2 x + 3cosx - 3 = 0\)

Thanks for your help in getting to the factorable equation.

Please would you check my answer?


After getting rid of the sin by using Pythagorus theorum we arrived at:

\(\displaystyle 2cos^2 x + 3cosx - 3 = 0\)

\(\displaystyle 2 cos ^2(x)-3 cos(x) + 1 = 0\)
\(\displaystyle (2 cos(x)-1) (cos(x)-1)\)
When
\(\displaystyle 2 cos(x)-1=0\)
\(\displaystyle cos(x)=1/2\)
\(\displaystyle x = pi/3\)
when
\(\displaystyle cos(x)-1 = 0\)
\(\displaystyle cos (x) = 1\)
\(\displaystyle x = 0\)


So the general rule is:

\(\displaystyle x = 2pi.n +/-( pi/3), 2.pi.n\)
 
I think you're forgetting the other solutions within the 0-to-2pi basic interval. These will give you more solutions in your "general" solution.

For instance, is cosine equal to zero only at x = 0...?

Eliz.
 
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