## anonymous 3 years ago PreCal Help Please

1. anonymous

Solve$4\sin^2x+4\sqrt{2} cosx-6=0$

2. amoodarya

the question is not here ! or i do not see it

3. anonymous

I just posted it before you wrote your comment

4. hba

Substitute in for the value of sin^2(x) You know what it is?

6. anonymous

Substitute it where? I am still confused.

sin^2(x) can be put in terms of cos^2(x)

sin^2(x) = 1-cos^2 (x) Another really important one to remember. (I hope you're writing these important ones down, especially if you plan on going further in calculus)

9. hartnn

then put cos x =y and you get a Quadratic in 'y', which can be solved easily.

10. hba

^ this sounds better

I wasn't going to suggest that.......but it makes it a lot easier. Go that route.

Don't forget to distribute when you substitute for sin^2(x)

13. anonymous

wait so is the new equation$4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx-6=0$

4(1-cos^2x).......

15. hartnn

actually, $\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x}$

16. amoodarya

no |dw:1358282546908:dw|

17. hartnn

$\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x \\ \implies 4\sin^2x=4-4\cos^2x}$

18. amoodarya

|dw:1358282634437:dw|

19. amoodarya

simplify that

20. anonymous

Can you rewrite that? It is hard to read

21. hartnn

$4\sin^2x+4\sqrt{2} cosx-6=0 \\ 4-4\cos^2x+4\sqrt{2} cosx-6=0\\4\cos^2x-4\sqrt{2} cosx+2=0 \\put \cos x=y \\ \implies 4y^2-4\sqrt2y+2=0$ can you solve this quadratic ?

That should be a -2 @hartnn

23. hartnn

u sure ?

24. anonymous

ok so the answer is $\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k$

25. hartnn

how did u get that ? what values of y u got ?

26. amoodarya

|dw:1358282921662:dw|

@hartnn 4-6=-2

28. hartnn

@Rosy95 did you actually solve it ?

29. amoodarya

|dw:1358283161601:dw|

30. anonymous

Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.

31. hartnn

@Rosy95 what answer you wrote is correct. did you solve it by yourself ? and any doubts anywhere ?

32. anonymous

I solved it by myself but i looked at the answer choices when I got half of it.

33. anonymous

Thanks everyone!

34. hartnn

ok, welcome ^_^