1. anonymous

let $$\large y=cos(2x)$$ so that trig equation can be written as: $$\large 2y^2=-1+3y$$ or $$\large 2y^2-3y+1=0$$ can you solve for y?

2. anonymous

so i plug the value for y into one of the equations?

3. anonymous

you realize that the trig equation you have is actually a trig equation in quadratic form??? so you would solve the quadratic equation by factoring: $$\large 2y^2-3y+1=0$$ $$\large (2y-1)(y-1)=0$$ so y = ????

4. anonymous

but its $2\cos ^{2}2\theta=-1+3\cos 2\theta$

5. anonymous

n dik wat 2 do w/the 2 in front of the theta

6. anonymous

read what i said... the trigonometric equation you have is a trig equation in QUADRATIC form.... to solve that trig equation, you would use the technique used to solve the quadratic equation: $$\large 2y^2=-1+3y$$

7. anonymous

so if i use the quadratic formula then i shud b able to solve for y?

8. anonymous

ok... this is your equation: $$\large 2\cdot \color {red}{cos^22\theta}=-1+3\cdot \color {red}{cos2\theta}$$ $$\large 2\cdot \color {red}{y^2}=-1+3\cdot \color {red}{y}$$ sure you can use the quadratic formula but it would be easier to factor as i did earlier...

9. anonymous

is y $(-3\pm sort{17})/4$

10. anonymous

no... rewrite the quadratic in standard form, then factor: $$\large 2y^2=-1+3y$$ $$\large 2y^2-3y+1=0$$ standard form $$\large (2y-1)(y-1)=0$$ factor so $$\large 2y-1=0$$ or $$\large y-1=0$$ now it's obvious y=1/2 , y = 1....

11. anonymous

n y stands for cos2theta?

12. anonymous

now that you have y=1/2 or y=1, remember that we replaced cos22θ=y so now you have to solve these two equations: $$\large cos2\theta=\frac{1}{2}$$ ; $$\large cos2\theta=1$$

13. anonymous

yes... y=cos2theta...

14. anonymous

so its 30

15. anonymous

n 0

16. anonymous

you have several answers between [0, 2pi)

17. anonymous

so 30 150 210 n 330?

18. anonymous

or just is it just 30 n 330 since we are dealing with cos?

19. anonymous

i only see 30, 330, 0 degrees

20. anonymous

well, the 30 and 330 degrees came from the first equation: $$\large cos2\theta=\frac{1}{2}$$ and the 0 degrees came from cos2theta = 1

21. anonymous

thank u!!!!!!! u helped so much.....by any chance cud u help me solve this next one? 6tanθ−6cotθ=0

22. anonymous

yw... post it up as a new question...

23. anonymous

okie dokie :)