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PreCal Help Please

Mathematics
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Solve\[4\sin^2x+4\sqrt{2} cosx-6=0\]
the question is not here ! or i do not see it
I just posted it before you wrote your comment

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Other answers:

  • hba
@amoodarya Refresh your page
Substitute in for the value of sin^2(x) You know what it is?
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)
then put cos x =y and you get a Quadratic in 'y', which can be solved easily.
  • 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)
wait so is the new equation\[4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx-6=0\]
4(1-cos^2x).......
actually, \[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x}\]
no |dw:1358282546908:dw|
\[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x \\ \implies 4\sin^2x=4-4\cos^2x}\]
|dw:1358282634437:dw|
simplify that
Can you rewrite that? It is hard to read
\[4\sin^2x+4\sqrt{2} cosx-6=0 \\ 4-4\cos^2x+4\sqrt{2} cosx-6=0\\4\cos^2x-4\sqrt{2} cosx+2=0 \\put \cos x=y \\ \implies 4y^2-4\sqrt2y+2=0\] can you solve this quadratic ?
That should be a -2 @hartnn
u sure ?
ok so the answer is \[\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k\]
how did u get that ? what values of y u got ?
|dw:1358282921662:dw|
@hartnn 4-6=-2
@Rosy95 did you actually solve it ?
|dw:1358283161601:dw|
Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.
@Rosy95 what answer you wrote is correct. did you solve it by yourself ? and any doubts anywhere ?
I solved it by myself but i looked at the answer choices when I got half of it.
Thanks everyone!
ok, welcome ^_^

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