Find all real numbers in the interval [0,2π) that satisfy the equation.?

Question

A. 2 cos (2x) - sqrt(3) = 0 

B. 3 sin x + 3 = 2 cos^2 x

Answer 1

(A). The equation is 2 cos(2x) - √3 = 0 and interval is [ 0, 2π ).

cos(2x) = √3/2

Let 2x = t, the function cos(t) has a period of 2π, first find all solutions in the interval [ 0, 2π ).

cos(t) = √3/2

cos(t) = cos(π/6).

The genaral solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer. 

t = 2nπ ± π/6

2x = 2nπ ± π/6

x = nπ ± π/12

x = π(n ± 1/12)

x = π(n - 1/12) or x = π(n + 1/12).

The solutions outside the interval [ 0, 2π ) are

If n = 0 then x = π[ (0) - 1/12 ] = - π/12 < 0

If n = 2 then x = π[ (2) + 1/12 ] = 25π/12 > 0.

The solutions in the interval [ 0, 2π ) are

x = π[ (0) + 1/12 ] = π/12,

x = π[ (1) - 1/12 ] = 11π/12,

x = π[ (1) + 1/12 ] = 13π/12,

x = π[ (2) - 1/12 ] = 23π/12.

The solutions in the interval [ 0, 2π ) are π/12, 11π/12, 13π/12 and 23π/12.

Answer 2

A). The trigonometric equation is .

The function cos(x) has a period of 2π, first find all solutions in the interval [0, 2π).

The function cos(x) is positive in first , fourth quadrant.

In first Quadrant, 0 ≤ x ≤ π/2.

In fourth Quadrant, 3π/2 ≤ x ≤ 2π.

, and

So the solutions are

.

, and

.

The solutions are in the interval [0, 2π).

Answer 3

B). The trigonometric equation is 3sin (x) + 3 = 2cos^2 (x).

Rewrite the equation as 3sin (x) + 3 = 2(1 - sin^2 (x))

3sin (x) + 3 = 2 - 2sin^2 (x)

2sin^2 (x) + 3sin (x) + 1 = 0

2sin^2 (x) + 2sin (x) + sin (x) + 1 = 0

2sin (x)[ sin (x) + 1 ] + 1[ sin (x) + 1 ] = 0

[ sin (x) + 1 ][ 2sin (x) + 1 ] = 0.

sin (x) + 1 = 0 or 2sin (x) + 1 = 0

sin (x) = - 1  or sin (x) = - 1/2

x = sin^{- 1}( - 1 ) or x = sin^{- 1}( - 1/2).

The function sin(x) has a period of 2π, first find all solutions in the interval [0, 2π).

The function sin(x) is negative in third and fourth quadrant.

In third Quadrant, π ≤ x ≤ 3π/2.

- 1 = - sin (π/2) = sin (π + π/2) = sin (3π/2).

- 1/2 = - sin (π/6) = sin (π + π/6) = sin (7π/6).

In fourth Quadrant, 3π/2 ≤ x ≤ 2π.

- 1/2 = - sin (π/6) = sin (2π - π/6) = sin (11π/6).

So the solutions are

x = sin^{- 1}(- 1) -----> x = sin^-1[sin (3π/2)] -------> x = 3π/2,

x = sin^{- 1}(- 1/2) -----> x = sin^-1[sin (7π/6)] -------> x = 7π/6, and

x = sin^{- 1}(- 1/2) -----> x = sin^-1[sin (11π/6)] -------> x = 11π/6^.

The solutions x = 3π/2, x = 7π/6, and x = 11π/6 in the interval [ 0, 2π).