Question

PreCal Help Please

Answer 1

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

Answer 2

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

Answer 3

I just posted it before you wrote your comment

Answer 4

@amoodarya Refresh your page

Answer 5

Substitute in for the value of sin^2(x)

You know what it is?

Answer 6

Substitute it where? I am still confused.

Answer 7

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

Answer 8

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)

Answer 9

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

Answer 10

^ this sounds better

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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

Answer 16

no

Answer 17

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