Solve the equation : (tan x + sqrt3)(2 cosx + 1) = 0in the interval 0 < x < 2pi?

(tan x + sqrt3)(2 cosx + 1) =0

→ Either (tan x + √3)=0

or (2 cosx + 1) =0

When tan x + √3=0 , we have

tanx=− √3

and x= π−π/3 and π+π/3 i.e 2π/3, 4π/3

When 2 cosx + 1=0

→cosx=− 1/2

i.e x=π−π/3 and π+π/3 i.e 2π/3, 4π/3

Hence 0 < x < 2π

x=2π/3, 4π/3

a million- sin 2x = sin^x + cos^x - 2sin x cos x = (cosx - sin x)^2 (utilising sin ^ x+ cos ^x =a million, sin 2x = 2 sinx cos x and then (a-b)^2 = a^2 -2ab +b^2 now (cosx - sin x)^2= cos x - sin x so cos x - sinx = 0 or cos x - sin x = a million (utilising y^2 = y => y^2-y = 0 =>y(y-a million) = 0 cos x -sin x = 0=>tan x = a million so x = pi/4 or 5pi/4(1st and 0.33 quadrant) cos x - sin x = a million LHS is of the type a cos x + b sin x( a = a million and b = -a million) if a = r cos y and b = r sin y r = 2^(a million/2), y = -pi/4 LHS = r cos y cos x + y sin y sin x = r cos(x+y) = a million cos(x+y) = a million/r = a million/2^a million(a million/2) = cos pi/4 x +y = pi/4 or -pi/4 x = pi/4 + y or -pi/4 +y = pi/4-pi/4 or -pi/4 -pi/4 = 0 or -pi/2 =0 or 3pi/2 (upload 2pi to make efficient appropriate quadrant) so we've answer 0, pi/4,5pi/4 3pi/2

(tan x + sqrt3)(2 cosx + 1) = 0

Either (tan x + sqrt3) = 0 or (2 cosx + 1) = 0

tan x=-sqrt3 or cosx=-1/2

-x=pi/3 and 4pi/3

x=-pi/3 and -4pi/3 {tan -a=tan a}

Add 2pi to both to get into interval

x=5pi/3 and 2pi/3

Cos x =1/2 {cos -a=cos a}

x=pi/3 and 4pi/3

x=pi/3, 5pi/3, 2pi/3 and 4pi/3

tan x + sqrt(3) = 0

tan x = -sqrt(3)

x = 2pi/3 or 5pi/3 (in quad 2 and 4)

2cos x + 1 = 0

cos x = -1/2

x = 2pi/3 or 4pi/3 (in quad 2 and 3)

2cos(x) + 1 = 0

2cos(x) = -1

cos(x) = -1/2

x = 2pi/3 , 4pi/3

tan(x) + sqrt(3) = 0

tan(x) = -sqrt(3)

x = 2pi/3 , 5pi/3

x = 2pi/3 , 4pi/3 , 5pi/3