## anonymous 5 years ago sin²θ-cos²θ-cosθ=1. Find all solutions in the interval [0,2π).

1. anonymous

$sin^2(x)-cos^2(x)-cos(x)=1$ first rewrite in term of cosine alone, using the identity $sin^2(x)+cos^2(x)=1$ $1-cos^2(x)-cos^2(x)-cos(x)=1$ $1-2cos^2(x)-cos(x)=1$ $-2cos^2(x)-cos(x)=0$ $2cos^(x)+cos(x)=0$ $cos(x)(2cos(x)+1)=$ $cos(x) = 0$ or $2cos(x)+1=0$

2. anonymous

of course you still have to solve $cos(x) = 0$ $x=\frac{\pi}{2}$or $x=\frac{3\pi}{2}$

3. anonymous

$2cos(x)+1=0$ $cos(x)=-\frac{1}{2}$ $x=\frac{2\pi}{3}$ $x=\frac{4\pi}{3}$

4. anonymous

so, x=2π/3 and 4π/3?

5. anonymous

you have 4 solutions all together.

6. anonymous

oh, ok...also the π/2 and 3π/2.... thanks

7. anonymous

right. is it clear?

8. anonymous

it is, when someone shows you!

9. anonymous

mmm rly u have unfintly solutin here by adding the period of ur functin

10. anonymous

of course! idea is to write using one trig function, then you get a quadratic equation to solve. there are an infinite number of solutions, that is why the question restricts interval from $[0,2\pi)$

11. anonymous

i.e ur sol + 2n bay

12. anonymous

thanks 'satellite 73' . hard to know where to start.

13. anonymous

welcome!

14. anonymous

mm sry 'm jst dont care 2 the question

15. anonymous

it is, when someone shows you!

16. anonymous

oh, ok...also the π/2 and 3π/2.... thanks

17. anonymous

so, x=2π/3 and 4π/3?

18. anonymous

yes 4 total. two for the equation $cos(x)=0$ and two for $cos(x)=-\frac{1}{2}$