Trigonometric Equations

Quote:

Solve the equation bolow given that 0 < x < 2pi
cos2x + 1/2 = 0
(sqrt(3) tan x + 1)(2 sinx - sqrt(3)) = 0

for the 1st one

cos(2x) = 2cos ²(x)-1

cos(2x) + 1/2 = 0
2cos ²(x) - 1 + (1/2) = 0
2cos ²(x) - (1/2) = 0
2cos ²(x) = 1/2
cos ²(x) = 1/4
cos(x) = +/- sqrt(1/4)
cos(x) = +/- 1/2
x = pi/3, 2pi/3, 4pi/3, 5pi/3

for the second one

(sqrt(3) tan x + 1)(2 sinx - sqrt(3)) = 0

this is true if either of the factors is 0 so

(sqrt(3)tan(x)+1) = 0 OR (2sin(x)-sqrt(3)) = 0

(sqrt(3)tan(x)+1) = 0
tan(x) = -1/sqrt(3)
x = arctan(-1/sqrt(3)) = 5pi/6, 11pi/6

(2sin(x) - sqrt(3)) = 0
sin(x) = sqrt(3)/2
x = arcsin(sqrt(3)/2) = pi/3, 2pi/3

so x = pi/3, 5pi/6, 2pi/3, 11pi/6 all satisfy the second equation

Thank you

zethwolves:

Thank you

What nice manners zethwolves. Rom will appreciate that.
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