Solve each trig equation on the interval 0 <or= to x < 2pi.
(1) 4 cos^2 (x) - 3 = 0
(2) tan(x/2) = sqrt{3}
What exactly do they mean by solving the equation
"on the interval 0 <or= to x < 2pi"?
Hello magentarita,
Since trig functions are cyclic i.e $\displaystyle f(x)=f(x+h)$(Ex: $\displaystyle \sin(x)=\sin(x+2\pi)$) for some fixed value of h the above equations will have infinite solutions. When given an interval it reduces the number of solutions.
The first one can be written as the difference of squares and we can use the zero product property
$\displaystyle 4\cos(x)-3=0 \iff [2\cos(x)]^2-(\sqrt{3})^2=0 \iff [2\cos(x)-\sqrt{3}][2\cos(x)+\sqrt(3)=0$
setting each factor above equal to zero we can solve the equation
$\displaystyle 2\cos(x)-\sqrt{3}=0 \iff \cos(x)=\frac{\sqrt{3}}{2} \iff x=\frac{\pi}{6} \mbox{ or } x=\frac{11\pi}{6}$
And the other factor
$\displaystyle 2\cos(x)+\sqrt{3}=0 \iff \cos(x)=-\frac{\sqrt{3}}{2} \iff x=\frac{5\pi}{6} \mbox{ or } x=\frac{7\pi}{6}$
Good luck