Solve the equation on the interval [ 0, 2 pi] (tan x + ?3)(2cos x + 1) = 0
you have a typo in your first bracket, what is (tanx + ?3) supposed to be ?
set each of the brackets equal to zero, I will do the second one
2cosx +1 = 0
cosx = -1/2
so x must be in the second or third quadrant according to the CAST rule.
(the cos 60º=+1/2)
so x = 180-60 = 120º or 2pi/3 radians
or x = 180 + 60 = 240º or 4pi/3 radians
do the same thing with the first bracket after you fix the typo.
(tan x + sqrt 3)(2 cos x + 1)=0
so from the first bracket,
tan x = - √3
now isn't tan 60 = +√3 ?
can you take it from there following the method I used for the second bracket?
let me know what you got.
Thanks I'll let you know what I get
To solve the equation (tan x + √3)(2cos x + 1) = 0 on the interval [0, 2π], we need to find the values of x that satisfy the equation.
To do this, we'll first set each factor equal to zero and solve for x.
1. Setting tan x + √3 = 0:
Subtracting √3 from both sides: tan x = -√3
2. Setting 2cos x + 1 = 0:
Subtracting 1 from both sides: 2cos x = -1
Dividing by 2: cos x = -1/2
Now, let's find the values of x that satisfy each equation within the given interval.
1. Solving tan x = -√3:
In the interval [0, 2π], the solutions for tan x = -√3 are:
x₁ = 11π/6 and x₂ = 7π/6
2. Solving cos x = -1/2:
In the interval [0, 2π], the solutions for cos x = -1/2 are:
x₃ = 2π/3 and x₄ = 4π/3
Therefore, the solutions to the equation (tan x + √3)(2cos x + 1) = 0 on the interval [0, 2π] are:
x₁ = 11π/6, x₂ = 7π/6, x₃ = 2π/3, and x₄ = 4π/3.