Question

Find all solutions in the interval [0, 2pi). give exact values and show all algebra.
2cos^2 X +3sinX-3=0

Answer 1

trick is to rewrite

\[cos^2(x)= 1-sin^2(x)\]

Answer 2

then get a quadratic equation in \[sin(x)\]

Answer 3

\[2cos^2(x)+3sin(x)-3=0\]
\[2(1-sin^2(x))+3sin(x)-3=0\]
\[2-2sin^2(x)+3sin(x)-3=0\]
\[-2sin^2(x)+3sin(x)-1=0\]
\[2sin^2(x)-3sin(x)-1=0\]

Answer 4

just solve as a quadratic. if it is easier to see replace sin(x) by z to get
\[2z^2-3z+1=0\]
\[(2z-1)(z-1)=0\]
\[z=\frac{1}{2}\] or \[z=1\]

Answer 5

then replace z by sin(x) and solve for x
\[sin(x)=1\]
\[x=\frac{\pi}{2}\]

Answer 6

\[sin(x)=\frac{1}{2}\]
\[x=\frac{\pi}{6}\] or \[x=\frac{5\pi}{6}\]

Answer 7

ooops second answer is wrong sorry

Answer 8

\[sin(x)=\frac{1}{2}\]
\[x=\frac{11\pi}{6}\]

Answer 9

k?

Answer 10

so x=11pi/6?

Answer 11

there are two places on the interval from 0 to 2pi where sine is 1/2

Answer 12

one is at
\[\frac{\pi}{6}\] and the other is at
\[\frac{11\pi}{6}\] so all together you have 3 answers.
those two and also
\[\frac{\pi}{2}\] where sine is 1

Answer 13

ohh i see thank you!

Answer 14

welcome!

Answer 15

any more?

Answer 16

huh?

Answer 17

any more of these to do?

Answer 18

well i just put some up

Answer 19

Does anyone happen to know 2cos^2 x+sinx-1=0 solve for x on [0,2pi)

Answer 20

@satellite73