## mathslover Group Title find the general solution for : $$\sin^2 \theta -2 \cos \theta + \frac{1}{4} = 0$$ one year ago one year ago

1. Yahoo! Group Title

$Sin^2\theta = 1-\cos^2 \theta$

2. Yahoo! Group Title

3. mathslover Group Title

I got $$\theta = \cos^{-1} \frac{1}{2}$$

4. mathslover Group Title

theta = 60 degrees.

5. mathslover Group Title

but general solution???

6. terenzreignz Group Title

Just add 2k(360 degrees) where k is any integer, and there's your general solution :D

7. Yahoo! Group Title

$\cos \theta =\frac{ 1 }{ 2 }$ $Cosx=Cosy$ then $x=(2n \Pi)\pm y$

8. terenzreignz Group Title

wait, no, k(360 degrees) not 2k sorry

9. calculusfunctions Group Title

May I explain?

10. mathslover Group Title

11. Yahoo! Group Title

$x= (2n \Pi) \pm \frac{ \Pi }{ 3 }$

12. mathslover Group Title

I had got this : cos theta = 1/2 , please explain after this.

13. Yahoo! Group Title

There Will be Two value For Cos x @mathslover

14. mathslover Group Title

Yes but the negative one will NOT BE ACCEPTED..

15. mathslover Group Title

as cos x can NOT BE NEGATIVE @Yahoo!

16. Yahoo! Group Title

Why min value of cos x = -1 ....

17. terenzreignz Group Title

Must x be in degrees?

18. terenzreignz Group Title

theta rather

19. Yahoo! Group Title

$\cos 180 = -1$

20. calculusfunctions Group Title

$\sin ^{2}\theta -2\cos \theta +\frac{ 1 }{ 4 }=0$First multiply the equation by the least common denominator.

21. mathslover Group Title

Wait, there will be two values for cos theta , the other one is -5/2 but -5/2 , but since the minimum value of cos theta is -1 and hence it can not be -5/2 is this what we must think abt @Yahoo! ? (sorry for my above explanation as : cos theta can never be negative as I thought that we are talking about cos(-theta) = cos theta)

22. mathslover Group Title

cos(theta) $$\ne$$ -5/2

23. Yahoo! Group Title

Yup.....nw u r Correct

24. calculusfunctions Group Title

@mathslover Let me know when you're interested in learning a more efficient method.

25. mathslover Group Title

@calculusfunctions sir, thanks a lot , I think I got it now.

26. mathslover Group Title

Sorry but do you have any easier or another way?

27. calculusfunctions Group Title

Yes if you're interested.