What is the equation tan(x+(pi/6)) equal to?
As with your other question, the expression is not an EQUATION. Your teacher is a moron, unless it's you who's introducing the wrongful use of that word. The expression tan(x + pi/6) = = [tan(x) + tan(pi/6)] / [1 - tan(x) tan(pi/6) ] = [tan(x) + 1/sqrt(3)] / [ 1 - tan(x)/sqrt(3) ] After the numerator and denominator have been multiplied through by sqrt(3), you will get expression D.
The equation tan(x+π/6) is equal to _____.... PLEASE HELP ME!!!?
Use the definition of tangent to rewrite problem as rational expression. tan(x + π/6) = sin(x + π/6)/cos(x + π/6) The sine and cosine of sum and difference formulas are handy to have memorized. sin(A + B) = sinAcosB + cosAsinB cos(A + B) = cosAcosB - sinAsinB tan(x + π/6) = [sinxcos(π/6) + cosxsin(π/6)]/cosxcos(π/6) - sinxsin(π/6)] You should also have memorized sine and cosine of π/6, which is 30° tan(x + π/6) = [(√3/2)sinx + (1/2)cosx]/[(√3/2)cosx - (1/2)sinx] Factoring (1/2) from numerator and denominator: tan(x + π/6) = [√3sinx + cosx]/[√3cosx - sinx] Factoring cosx from numerator and denominator: tan(x + π/6) = [√3tanx + 1]/[√3 - tanx]
If x+y=π/2 and y+z=x, then tan x is equal to (A) 2 (tan y+tan z) (B) tan y+tan z (C) tan y +2tan z?
[math]x+y=\pi/2=》x=\pi/2-y[/math]. So, [math]tan x=tan(\pi/2-y)=cot y[/math] or [math](tan x)(tan y)=1[/math].Seeing the second equation y+z=x=》z=x-y.Taking tangent on both sides, [math]tan z=(tan x-tan y)/(1+(tan x)(tan y))[/math] or [math]tan z=(tan x-tan y)/(1+1)[/math] (since tan x tan y=1) or [math]2tan z=tan x-tan y[/math] which gives [math]tan x=tan y+2tan z[/math]Which gives the option (C).
Solve the equation (tanx+1)( 2sinx - sqrt(3) ) = 0?
The term on the left side will be zero whenever one of its factors is 0. The first factor tanx+1 is zero whenever tanx = -1. In the given interval, that's at x=3π/4 and at x=7π/4. The second factor 2sinx - √3 is zero whenever sinx = √3/2. In the given interval, that's at x=π/3 and at x = 2π/3. So the solution set of this equation is {π/3, 2π/3, 3π/4, 7π/4}
How do you find the solutions of tan (x-pi/4) =0, 0<=x<2pi?
Tan(45)=1Pi equal to 180180/4= 45Pi/4=45degreeTan(x-45 )=0Tan can be equal 0 just, in 0degree and 180 degreeTan can be 0 in 180nn= natural numbersBut in question want answer just in 360degree(X<2pi)Tan(180)=0 Tan(0)=0So we have two probability(x-45)=0 Or (x-45)=180X1= 45 X2= 225
Find general solution of the equation tan^2 x+ 2* (3^1/2) tan x= 1?
[math]\tan^2x+2\times3^\frac{1}{2}\tan x=1[/math][math]\implies\tan^2x+2\sqrt{3}\tan x-1=0[/math]This is a quadratic equation in [math]\tan x[/math].[math]\implies\tan x=\cfrac{-2\sqrt{3}\pm\sqrt{(2\sqrt{3})^2-4\times1\times(-1)}}{2\times1}[/math][math]\implies\tan x=\cfrac{-2\sqrt{3}\pm\sqrt{12+4}}{2}[/math][math]\implies\tan x=\cfrac{-2\sqrt{3}\pm\sqrt{16}}{2}[/math][math]\implies\tan x=\cfrac{-2\sqrt{3}\pm4}{2}[/math][math]\implies\tan x=-\sqrt{3}\pm2[/math]The principal solution would be the following.[math]\tan x=-\sqrt{3}-2\implies x=\cfrac{-5\pi}{12}[/math]or[math]\tan x=-\sqrt{3}+2\implies x=\cfrac{\pi}{12}[/math]The general solution is then the following set.[math]S=\left\lbrace x|x=m\pi-\cfrac{5\pi}{12},m\in\mathbb{Z}\right\rbrace\cup\left\lbrace x|x=n\pi+\cfrac{\pi}{12},n\in\mathbb{Z}\right\rbrace[/math]
What is the general solution of tan x+ cot x=2?
[math]tan(x) + cot(x) = \dfrac{sin(x)}{cos(x)} + \dfrac{cos(x)}{xin(x)} = \dfrac{sin^2(x) + cos^2(x)}{sin(x)cos(x)} =[/math][math]= \dfrac{2}{sin(2x)}[/math][math]\dfrac{2}{sin(2x)} = 2[/math][math]sin(2x) = 1[/math][math]2x = \dfrac{\pi}{2} + 2k\pi[/math][math]x= \dfrac{\pi}{4} +k\pi[/math]
What is the general solution of x for (tan x) * (tan2x) = 1?
We know: tan 2x = 2 tanx / (1 - tan^2x).Given equation is:tan x * 2* tan x / ( 1- tan ^2 x) = 12 tan^2x = 1- tan^2 x.tan x = sqrt (1/3)x = tan universe of sqrt (1/3)x = pi/6. {x is angle expressed in radians.}General solution is: x = 2* n * pi + pi/6.(where n is a natural number).Sanjay C.