Question

1. Solve for x in the interval [0 , 2 π ).
(a) 2sin^2x=sin x
(b) 3sin^2x+cos^2x=2
(c) 1+3cosx=cos2x
(d) sin2x+sin x=0

Answer 1

a) rearrange to get 2sin^2x-sinx=0
then factor sinx(2sinx-1)=0
then set both sinx and 2sinx-1 equal to 0 and solve
sinx=o means at what angle is the sin 0, this is 0
2sinx-1=0 rearrange to get sinx=1/2
at what angle is the sin 1/2, this is pi/6 and 5pi/6, I believe.

Answer 2

b)using the identity sin^2x +cos^2x=1 rearranged to cos^2x=1-sin^2x
3sin^2x+1-sin^2x=2
combining like terms we get
2sin^2x=1
solving for sinx we get sinx=+-sqrt(2)/2
At what angle is the sin +-sqrt(2)/2?
That would be pi/4, 3pi/4, 5pi/4 and 7pi/4

Answer 3

c) cos2x=2cos^2 x - 1
substituting in we get
1+3cosx=2cos^2x-1
Rearranging we get
2cos^2x-3cosx-2=0
Factoring we get
(2cosx+1)(cosx-2)=0
setting both = to 0 gives
cosx=-1/2 and cosx=2
Now Im not sure what to do about the cosx=2 part but what angle has a cos of -1/2?
That would be 2pi/3 and 4pi/3, I think.

Answer 4

d) using the double angle formula sin2x=2sinxcosx
we substitute and get
2sinxcosx+sinx=0
Factoring out a sinx we get
sinx(2cosx+1)=0
setting both = to 0 we get
sinx=0 and 2cosx+1=0
what angle in the interval has a sin of 0 and a cos of -1/2
this is 0, pi (I forgot this solution in part a), 2pi/3 and 4pi/3

Answer 5

Hope this helps. Please check the math as I did it rather quickly and somewhat distractedly :)

Answer 6

have cjecked it, perfect thanks :)