Simplifying Trigonometric Functions?

Question

I completely forgot how to simplify trigonometric functions. Here is a link to the question I am trying to answer:

http://img527.imageshack.us/img527/3470/picture1mx1.jpg

Thanks in advance for your help!

Answer 1

sin2x = 2sinxcosx

sin2x + cosx = 0

2sinxcosx + cosx = 0

cosx (2sinx + 1) = 0

cosx = 0    or    2sinx + 1 = 0

You should be able to finish it off.

Answer 2

Since sin2x = 2sinxcosx

sin2x + cosx = 0

(2sinxcosx) + cosx = 0

cosx (2sinx + 1) = 0

cosx = 0 or 2sinx + 1 = 0

x = pi/2 , 3pi/2, 7pi/6, 11pi/6,

THERE SHOULD BE FOUR SOLUTIONS within the domain [0,2pi] hence the above is missing a solution.

Answer 3

sin2X + cosX=0       (sin2X=2sinXcosX)

2sinXcosX + cosX=0 (I factor out the cosX)

cosX(2sinX+1)=0

so

cosX=0      or   2sinX+1=0

X=pi/2 or 3pi/2 or 2sinX= -1

                             sinX= -.5

                                 X= -pi/6, which is 11pi/6

the answer are Pi/2, 3Pi/2, 11pi/6