Apr 25, 2015 · How do you use inverse trigonometric functions to find the solutions of the equation that are in

Dec 2, 2017 · The answer is pi/6 We can do that easily by rationalizing the numerator we get tan^-1[(sqrt3)/3]=tan^-1 ...

Aug 5, 2017 · theta=pi/3+npi,n inZZ "given "tantheta=sqrt3 rArrtheta=tan^-1(sqrt3)=pi/3 "since the tangent function is periodic" "we can generalise the solution as" theta=pi/3+npi,n inZZ

Jun 13, 2018 · x = pi/3 + kpi x = pi/4 + kpi tan^2 x - (1 + sqrt3)tan x + sqrt3 = 0 Solve this quadratic equation for tan x. Use the improved quadratic formula (Socratic, Google Search) D = d^2 = b^2 - 4ac = (1 + sqrt3)^2 - 4sqrt3 = 4 - 2sqrt3 = 0.536 --> d = +- 0.73 There are 2 real roots: tan x = -b/(2a) +- d/(2a) = (1 + sqrt3)/2 +- 0.73/2 = 1.37 +- 0.37 tan x = 1.37 + 0.37 = 1.73 = sqrt3, and tan x = 1.37 ...

Aug 22, 2015 · Solve tan^2 x = 1/3 tan^2 x = 1/3 --> tan x = +- 1/sqrt3 = +- sqrt3/3 a. tan x = sqrt3/3 --> x = pi/6 + kpi b. tan x = - sqrt3/3 --> x = (5pi)/6 + kpi

Aug 6, 2016 · =-30 tan^-1(-1/sqrt3) =-30

Aug 23, 2015 · Solve tan^2 x - sqrt3tan x = 0 Put tan x in common factor: tan x(tan x - sqrt3) = 0 a. tan x = 0 --> x = 0 + kpi b. tan x = sqrt3 --> x = pi/3 + kpi

Jan 10, 2017 · Given #tanx=sqrt3/3, cosx=-sqrt3/2# to find the remaining trigonometric function? Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer

Aug 13, 2016 · sqrt3 Using unit circle and trig table of special arcs --> tan ((4pi)/3) = tan (pi/3 + pi) = tan (pi/3) = sqrt3

Dec 25, 2016 · If #tan theta=2/3# and #cos theta >#, how do you find #sin theta#? In what quadrants will an angle in standard position have a positive tangent value? How is #150^@# related to #30^@# on the unit circle?