## anonymous 5 years ago Find all solutions in the interval [0, 2pi). give exact values and show all algebra. 2cos^2 X +3sinX-3=0

1. anonymous

trick is to rewrite $cos^2(x)= 1-sin^2(x)$

2. anonymous

then get a quadratic equation in $sin(x)$

3. anonymous

$2cos^2(x)+3sin(x)-3=0$ $2(1-sin^2(x))+3sin(x)-3=0$ $2-2sin^2(x)+3sin(x)-3=0$ $-2sin^2(x)+3sin(x)-1=0$ $2sin^2(x)-3sin(x)-1=0$

4. anonymous

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 $z=1$

5. anonymous

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

6. anonymous

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

7. anonymous

ooops second answer is wrong sorry

8. anonymous

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

9. anonymous

k?

10. anonymous

so x=11pi/6?

11. anonymous

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

12. anonymous

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

13. anonymous

ohh i see thank you!

14. anonymous

welcome!

15. anonymous

any more?

16. anonymous

huh?

17. anonymous

any more of these to do?

18. anonymous

well i just put some up

19. anonymous

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

20. anonymous

@satellite73