Question

Need a bit of help with trigonometry equations.

Answer 1

\[2 \sin^{2} x + 15 \cos x -9 =0 \] if \[0 \le x, \theta \le 2\pi\]

Answer 2

you can write this in just in terms of cos(x)
using the Pythagorean identity
sin^2(x)=1-cos^2(x)

this will result in a quadratic in terms of the function cos(x)

Answer 3

So I can replace cox x with another term such as say, "u" and solve accordingly?

Answer 4

you can if you want yep

Answer 5

then once you simplified this thing you called u
you will have to replace u with cos(x)
and solve for x in the given interval

Answer 6

assume a not zero
\[a(\cos(x))^2+b(\cos(x))+c=0 \\ \cos(x)=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
or you don't have to use quadratic formula
if the thing you have is actually factorable :)

Answer 7

Oh I see. :O Wow thanks!

Answer 8

\[\frac{ \pi }{ 3}, \frac{ 5\pi }{ 3 } \]

Answer 9

one sec while I check

Answer 10

\[2\sin^2(x)+15\cos(x)-9=0 \\ 2(1-\cos^2(x))+15\cos(x)-9=0 \\ 2-2\cos^2(x)+15\cos(x)-9=0 \\ -2\cos^2(x)+15\cos(x)-7=0 \\ -2\cos^2(x)+14\cos(x)+\cos(x)-7=0 \\ -2\cos(x)(\cos(x)-7)+1(\cos(x)-7)=0 \\ (\cos(x)-7)(-2\cos(x)+1)=0\]
cos(x)-7=0 can't happen
so you have cos(x)=-1/2

Answer 11

oops cos(x)=1/2

Answer 12

you are right those are the only solutions in the interval [0,2pi]

Answer 13

notice I didn't use quadratic formula
I used factoring :)