# Find the point of intersections of he curves defined by y = sin 2x and y = tan x for 0≤x≤2pi.

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### 1 Answer

The point of intersection of the curves y = sin 2x and y = tan x for values of x that satisfy 0≤x≤2pi have to be determined.

y = sin 2x = 2*sin x*cos x

y = tan x = `(sin x)/cos x`

2*sin x*cos x = `sin x /cos x`

The two are equal when sin x = 0

=> x = `sin^-1(0)`

=> x = 0 and x = `pi` and x = `2*pi`

Canceling sin x gives

=> 2*cos x = `1/cos x`

=> `cos^2 x = 1/2`

=> cos x = `1/sqrt 2` and cos x = `-1/sqrt 2`

=> x = `cos^-1(1/sqrt 2)` and x = `cos^-1(-1/sqrt 2)`

For 0 ≤ x ≤ 2pi

x = `pi/4 ` and `2pi - pi/4 = (7*pi)/8 ` and x = `pi - pi/4 = (3*pi)/4` and x = `pi + pi/4 = (5*pi)/4`

**The points of intersection of y = sin 2x and y = tan x for values of x that satisfy 0 ≤ x ≤ 2pi are `(0,0),(pi, 0),(2*pi, 0)(pi/4, 1/sqrt 2), ((7*pi)/4, 1/sqrt 2), ((3*pi)/4, -1/sqrt 2) and ((5*pi)/4, -1/sqrt 2)` **