The following equation involves multiple angles. Solve the equation on the interval (0, 2pi) tan x/2 = - suare root 3\ 3

Step-by-step explanation:

tan x/2=-√3/3

sec²x/2-tan²x/2=1

sec²x/2-3/9=1

sec² x/2=1+1/3=4/3

cos² x/2=3/4

2cos² x/2=3/2

1+cos x=3/2

cos x=3/2-1=1/2=cos π/3,cos (2π-π/3)

x=π/3,5π/3

or

cos x=(1-tan²x/2)/(1+tan²x/2)=(1-1/3)(1+1/3)=(2/3)/(4/3)=1/2=cos π/3,cos (2π-π/3)

or x=π/3,5π/3

x=π/3

x/2=π/6

tan x/2=tan π/6=√3/3

so x=π/3 is an extraneous value.

x=5π/3 is the solution.

or

tan x/2=-√3/3=-1/√3=(-1/2)/(√3/2) [by dividing numerator and denominator by 2]

=-tan π/6=tan (π-π/6),tan (2π-π/6)

x/2=5π/6,11π/6

x=5π/3,11π/3,

now 11π/3 >2π

so x=5π/3

answer: i don't understand it

step-by-step explanation: sorry