Question

how do i solve the equation on the interval
0 le theta < 2pi.

cos(2 theta)-3 cos theta + 2 = 0

Answer 1

anybody can helppp???

Answer 2

nobody there???

Answer 3

cos (2 theta) = 2 cos ^2 (theta) -1
replace and treat cos (theta ) as x , solve for x, then, plug back to solve for theta.

Answer 4

\[\cos 2\theta-3\cos \theta+2=0,2\cos ^{2}\theta-1-3\cos \theta+2=0\]
\[2\cos ^{2}\theta-3\cos \theta+1=0\]
\[2\cos ^{2}\theta-2\cos \theta-\cos \theta+1=0,\]
\[2\cos \theta \left( \cos \theta-1 \right)\]
complete it/

Answer 5

answer has to be one of these:
a) \[0, \pi/6, \pi, 11\pi/6\]
b) \[\pi/3, \pi/2, 5\pi/3\]
c) \[0, \pi/3, 5\pi/3\]
d) 0, \[\pi/6, 11\pi/6\]

Answer 6

Can someone help with this???

Answer 7

\[2\cos \theta \left( \cos \theta-1 \right)-1\left( \cos \theta-1 \right)=0\]
\[\left( \cos \theta-1 \right)\left( 2\cos \theta-1 \right)=0\]
\[either \cos \theta-1=0,\cos \theta=1=\cos 0, \in 0\le \theta < 2\pi ,\theta=0 \]
\[or 2\cos \theta-1=0,\cos \theta=\frac{ 1 }{2 }=\cos \frac{ \pi }{ 3 },\cos \left( 2\pi-\frac{ \pi }{ 3 } \right)\]
\[\theta=\frac{ \pi }{ 3 },\frac{ 5\pi }{ 3 }\]

Answer 8

Thanks

Answer 9

yw