Rosy95
PreCal Help Please
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Rosy95
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Solve\[4\sin^2x+4\sqrt{2} cosx-6=0\]
amoodarya
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the question is not here !
or i do not see it
Rosy95
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I just posted it before you wrote your comment
hba
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@amoodarya Refresh your page
cgreenwade2000
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Substitute in for the value of sin^2(x)
You know what it is?
Rosy95
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Substitute it where? I am still confused.
cgreenwade2000
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sin^2(x) can be put in terms of cos^2(x)
cgreenwade2000
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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)
hartnn
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then put cos x =y
and you get a Quadratic in 'y', which can be solved easily.
hba
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^ this sounds better
cgreenwade2000
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I wasn't going to suggest that.......but it makes it a lot easier. Go that route.
cgreenwade2000
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Don't forget to distribute when you substitute for sin^2(x)
Rosy95
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wait so is the new equation\[4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx-6=0\]
cgreenwade2000
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4(1-cos^2x).......
hartnn
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actually, \[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x}\]
amoodarya
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no
|dw:1358282546908:dw|
hartnn
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\[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1-\cos^2x \\ \implies 4\sin^2x=4-4\cos^2x}\]
amoodarya
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|dw:1358282634437:dw|
amoodarya
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simplify that
Rosy95
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Can you rewrite that? It is hard to read
hartnn
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\[4\sin^2x+4\sqrt{2} cosx-6=0 \\ 4-4\cos^2x+4\sqrt{2} cosx-6=0\\4\cos^2x-4\sqrt{2} cosx+2=0 \\put \cos x=y \\ \implies 4y^2-4\sqrt2y+2=0\]
can you solve this quadratic ?
cgreenwade2000
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That should be a -2 @hartnn
hartnn
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u sure ?
Rosy95
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ok so the answer is \[\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k\]
hartnn
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how did u get that ?
what values of y u got ?
amoodarya
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|dw:1358282921662:dw|
cgreenwade2000
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@hartnn 4-6=-2
hartnn
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@Rosy95 did you actually solve it ?
amoodarya
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|dw:1358283161601:dw|
Rosy95
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Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.
hartnn
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@Rosy95 what answer you wrote is correct.
did you solve it by yourself ? and any doubts anywhere ?
Rosy95
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I solved it by myself but i looked at the answer choices when I got half of it.
Rosy95
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Thanks everyone!
hartnn
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ok, welcome ^_^