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

2sin^2x-sinx=0
sinx(2sinx-1)=0
sinx=0 or 2sinx-1=0
if sinx = 0, x = 0, pi

If 2sinx-1=0, sinx = 1/2 and x =
pi/6, 5pi/6

Answer 2

\[2\sin^2(x)=\sin(x)\]
\[2\sin^2(x)-\sin(x)=0\]
\[\sin(x)(2\sin(x)-1)=0\]
\[\sin(x)=0, \sin(x)=\frac{1}{2}\]
\[x=0, x =\pi\] for the first one
\[x=\frac{\pi}{6}, x=\frac{5\pi}{6}\] for the second

Answer 3

what mertsj said

Answer 4

d)
\[\sin(2x)+\sin(x)=0\]
\[2\sin(x)\cos(x)+\sin(x)=0\]
\[\sin(x)(2\cos(x)+1)=0\]
\[\sin(x)=0, \cos(x)=-\frac{1}{2}\]
\[x=0, x=\frac{2\pi}{3}, x=\frac{4\pi}{3}\]

Answer 5

(b) replace sin^2x with 1-cos^2x and get 3(1-cos^x)+cos^2x = 2 or 3-3cos^2x+cos^2x=2
-2cos^2=-1
cos^2x = 1/2
cos x = pos or neg sqrt2/2
x = pi/4,3pi/4, 5pi/4, 7pi/4

Answer 6

mertsj am not getting the same interval with u