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Rosy95
 3 years ago
PreCal Help Please
Rosy95
 3 years ago
PreCal Help Please

This Question is Closed

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Solve\[4\sin^2x+4\sqrt{2} cosx6=0\]

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0the question is not here ! or i do not see it

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0I just posted it before you wrote your comment

hba
 3 years ago
Best ResponseYou've already chosen the best response.0@amoodarya Refresh your page

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Substitute in for the value of sin^2(x) You know what it is?

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Substitute it where? I am still confused.

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1sin^2(x) can be put in terms of cos^2(x)

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1sin^2(x) = 1cos^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
 3 years ago
Best ResponseYou've already chosen the best response.1then put cos x =y and you get a Quadratic in 'y', which can be solved easily.

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1I wasn't going to suggest that.......but it makes it a lot easier. Go that route.

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1Don't forget to distribute when you substitute for sin^2(x)

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0wait so is the new equation\[4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx6=0\]

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.14(1cos^2x).......

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1actually, \[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1\cos^2x}\]

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0no dw:1358282546908:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\[\huge \color{red}{\sin^2x+\cos^2x=1 \\ \implies \sin^2x =1\cos^2x \\ \implies 4\sin^2x=44\cos^2x}\]

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358282634437:dw

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Can you rewrite that? It is hard to read

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1\[4\sin^2x+4\sqrt{2} cosx6=0 \\ 44\cos^2x+4\sqrt{2} cosx6=0\\4\cos^2x4\sqrt{2} cosx+2=0 \\put \cos x=y \\ \implies 4y^24\sqrt2y+2=0\] can you solve this quadratic ?

cgreenwade2000
 3 years ago
Best ResponseYou've already chosen the best response.1That should be a 2 @hartnn

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0ok so the answer is \[\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1how did u get that ? what values of y u got ?

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358282921662:dw

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1@Rosy95 did you actually solve it ?

amoodarya
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1358283161601:dw

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1@Rosy95 what answer you wrote is correct. did you solve it by yourself ? and any doubts anywhere ?

Rosy95
 3 years ago
Best ResponseYou've already chosen the best response.0I solved it by myself but i looked at the answer choices when I got half of it.
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