# trig

• Jun 13th 2008, 04:30 PM
meli3000
trig
solve the following system of equations algebraically (-2π, 2π)

y=cos^2 (x)
y=sin^2 (x) + 1/2

Thanks
• Jun 13th 2008, 04:34 PM
sean.1986
Cos^2(x) = Sin^2(x) + 1/2

Cos^2(x) - Sin^2(x) = 1/2

Cos(2x) = 1/2

2x = Cos^-1(1/2)

x = (Cos^-1(1/2))/2
• Jun 13th 2008, 04:52 PM
sean.1986
In fact, it's...

pi/6, 5pi/6, -pi/6, -5pi/6
• Jun 13th 2008, 07:06 PM
Soroban
Hello, meli3000!

Quote:

Solve the following system of equations algebraically $(-2\pi,\:2\pi)$

. . ${\color{blue}[1]}\;\;y \;=\;\cos^2\!x$
. . ${\color{blue}[2]}\;\;y\:=\:\sin^2\!x + \frac{1}{2}$

$\text{Add {\color{blue}[1]} and {\color{blue}[2]}: }\;2y \;=\;\underbrace{\cos^2\!x+\sin^2\!x}_{\text{This is 1}} + \frac{1}{2} \quad\Rightarrow\quad 2y \:=\:\frac{3}{2} \quad\Rightarrow\quad\boxed{y \:=\:\frac{3}{4}}$

Substitute into [1]: . $\cos^2\!x \:=\:\frac{3}{4}\quad\Rightarrow\quad \cos x \:=\:\pm \frac{\sqrt{3}}{2}$

. . $\boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\pm\fr ac{7\pi}{6},\:\pm\frac{11\pi}{6}}$

• Jun 13th 2008, 07:12 PM
sean.1986
Ah yeah forgot that because it's double you get 2 more answers. Oops!