Can someone please help? tan(x/2)=(sqrt3)/3
\[\large\rm 0.5x= tan^{-1}\frac{√3}{3} \]
But for what intervals are you finding the values of x?
lt doesnt say the options are x=2pi/3 +pi.....x=pi/3+2npi......x=7pi/6+npi and 11pi/6 + npi....x=5pi/3+npi......x=5pi/6+npi and 7pi/6+npi..........All where n is an integer
i need help bro
@freckles please
please
i need help
\[\tan(\frac{x}{2})=\frac{\sqrt{3}}{3} \\ \text{ \let }\theta=\frac{x}{2} \\ \text{ do you know how to solve } \\ \tan(\theta)=\frac{\sqrt{3}}{3} \\ \text{ also know as } \\ \tan(\theta)=\frac{1}{\sqrt{3}}\]
whip out your unit circle and find when sin is 1/2 and cos is sqrt(3)/2 find also when sin is -1/2 and cos is -sqrt(3)/2
once we find theta we will go back and find x
pi/6 and 5pi/6.....7pi/6 and 11pi/6
well at 11pi/6 and 5pi/6 you have tan is negative this is why I asked you to look only at the positive, positive quadrant and the negative, negative quadrant that is the ordered pairs (sqrt(3)/2,1/2) and (-sqrt(3)/2,-1/2) belong to those quadrants I just mentioned
so we will use \[\theta=\frac{\pi}{6} +2n \pi \\ \text{ or } \\ \theta=\frac{7\pi}{6}+2n \pi\]
recall the equation to solve want tan(x/2)=sqrt(3)/3 but we replaced x/2 with theta
so you have to solve the following two equations for x: \[\frac{x}{2}=\frac{\pi}{6}+2n \pi \\ \frac{x}{2}=\frac{7\pi}{6}+2n \pi\]
hmm
never learned that. how do u solve it?
It is basic fractions
pi/3?
sorry it is super late
so you never solved equations like x/2=6 ?
you multiply both sides by ...
2
also I could have condensed my equations into one equation
yes
@freckles Are you sure the equations to which you simplified are correct since there's no mention of the answer derived through them in the choices
my answer is correct however I just mentioned that I could condense my two equations into one equation
\[\frac{x}{2}=\frac{\pi}{6}+n \pi\] since the line segment from (sqrt(3)/2,1/2) and (-sqrt(3)/2,-1/2) has an angle of 180 degrees
anyways yes you just must multiply both sides by 2
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