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Rosy95Best ResponseYou've already chosen the best response.0
Solve\[4\sin^2x+4\sqrt{2} cosx6=0\]
 one year ago

amoodaryaBest ResponseYou've already chosen the best response.0
the question is not here ! or i do not see it
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
I just posted it before you wrote your comment
 one year ago

hbaBest ResponseYou've already chosen the best response.0
@amoodarya Refresh your page
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
Substitute in for the value of sin^2(x) You know what it is?
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
Substitute it where? I am still confused.
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
sin^2(x) can be put in terms of cos^2(x)
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
sin^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)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
then put cos x =y and you get a Quadratic in 'y', which can be solved easily.
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
I wasn't going to suggest that.......but it makes it a lot easier. Go that route.
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
Don't forget to distribute when you substitute for sin^2(x)
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
wait so is the new equation\[4( \frac{ 1 }{\cos^2x }) + 4 \sqrt{2}cosx6=0\]
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
4(1cos^2x).......
 one year ago

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

amoodaryaBest ResponseYou've already chosen the best response.0
no dw:1358282546908:dw
 one year ago

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

amoodaryaBest ResponseYou've already chosen the best response.0
dw:1358282634437:dw
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
Can you rewrite that? It is hard to read
 one year ago

hartnnBest 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 ?
 one year ago

cgreenwade2000Best ResponseYou've already chosen the best response.1
That should be a 2 @hartnn
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
ok so the answer is \[\frac{ \pi }{4 }+2\pi k, \frac{ 7\pi }{4 }+2\pi k\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
how did u get that ? what values of y u got ?
 one year ago

amoodaryaBest ResponseYou've already chosen the best response.0
dw:1358282921662:dw
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
@Rosy95 did you actually solve it ?
 one year ago

amoodaryaBest ResponseYou've already chosen the best response.0
dw:1358283161601:dw
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
Im' sorry, I forgot to include the rest of the question...it is solve (that long equation) for all real values of x.
 one year ago

hartnnBest 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 ?
 one year ago

Rosy95Best ResponseYou've already chosen the best response.0
I solved it by myself but i looked at the answer choices when I got half of it.
 one year ago
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