Finding all Solutions for

2cos(x)-�ã3=0

5 answers

Is this cosx=1/2sqrt3 ?
no it's 2cosx-sqrt3=0
I will read it as
2cosx - √3 = 0
cosx = √3/2
x = 30° , from the standard 30-60-90 triangle

or x = 330°

"all solutions " would be
x = 30 + k(360) or x = 330+k(360)° where k is an integer.
How do you find the solutions?
since the standard cosine curve has a period of 360°, any answer you have will repeat itself in the next curve or 360° either to the right or to the left.
by multiplying 360 by k, k and integer, I am simply adding/or subtracting multiples of 360 to any answer.

to get my original 30°, as I said, I recognized the rations of the 30-60-90 right-angled triangle, which are 1:√3:2.
Secondly , since the cosine is positive in the 1st and 4th quadrant, the 30° will be the 1st quadrant solution and 360-30 or 330° would be the 4th quadrant solution.
Similar Questions
  1. Find all solutions on the interval [0.2pi)A) -3sin(t)=15cos(t)sin(t) I have no clue... b) 8cos^2(t)=3-2cos(t) All i did was move
    1. answers icon 1 answer
  2. Given that 2cos^2 x - 3sin x -3 = 0, show that 2sin^2 x + 3sin x +1 = 0?Hence solve 2cos^2 x + 4sin x - 3 = 0, giving all
    1. answers icon 2 answers
  3. cosA= 5/9 find cos1/2Aare you familiar with the half-angle formulas? the one I would use here is cos A = 2cos^2 (1/2)A - 1 5/9 +
    1. answers icon 0 answers
  4. I am given a vector function <2cos(t) + cos(2t), 2sin(t) + sin(2t), 0>(in other words) : (2cos(t) + cos(2t))i + (2sin(t) +
    1. answers icon 2 answers
more similar questions