Trigonometry help!

Question

Answer 1

• 13).

The trigonometric equation is 4sin² x + 4cos x - 5 = 0.

Pythagorean identity : sin² x + cos² x = 1.

4(1 - cos² x) + 4cos x - 5 = 0

4 - 4cos² x + 4cos x - 5 = 0

4cos² x - 4cos x + 1 = 0

4cos² x - 2cos x - 2cos x + 1 = 0

2cos x(2cos x - 1) - 1(2cos x - 1) = 0

(2cos x - 1)² = 0

2cos x - 1 = 0

⇒ cos x = 1/2.

cos x = cos (π/3).

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

⇒x = 2nπ ± (π/3).

If n = 0, x = 2(0)π + (π/3) and x = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, x = 2(1)π + (π/3) and x = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

Therefore, the solutions of the given equation are x = π/3 and x = 5π/3 in the interval [0, 2π).

• 14). The trigonometric equation is 2cos² (2x) + 3cos (2x) + 1 = 0.

Let, cos(2x) = t.

Then, the equation is 2t² + 3t + 1 = 0.

2t² + 2t + t + 1 = 0

2t(t + 1) + 1(t + 1) = 0

(t + 1)(2t + 1) = 0

⇒t + 1 = 0 and 2t + 1 = 0

t = - 1 and t = - 1/2

Put, t = cos(2x).

cos (2x) = - 1 and cos (2x) = - 1/2.

• cos (2x) = - 1.

cos (2x) = cos(π)

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

2x = 2nπ ± π

⇒x = nπ ± (π/2)

If n = 0, x = (0)π + (π/2) and x = (0)π - (π/2) = π/2 and - π/2.

If n = 1, x = (1)π + (π/2) and x = (1)π - (π/2) = π + π/2 and π - π/2 = 3π/2 and π/2,

If n = 2, x = (2)π + (π/2) and x = (2)π - (π/2) = 2π + π/2 and 2π - π/2 = 5π/2 and 3π/2.

Therefore, the solutions of the given equation are x = π/2 and x = 3π/2 in the interval [0, 2π).

• cos (2x) = - 1/2.

cos (2x) = cos(2π/3)

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

2x = 2nπ ± (2π/3)

⇒x = nπ ± (π/3)

If n = 0, x = (0)π + (π/3) and x = (0)π - (π/3) = π/3 and - π/3.

If n = 1, x = (1)π + (π/3) and x = (1)π - (π/3) = π + π/3 and π - π/3 = 4π/3 and 2π/3,

If n = 2, x = (2)π + (π/3) and x = (2)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

Therefore, the solutions of the given equation are x = π/3, x = 2π/3, x = 4π/3 and x = 5π/3 in the interval [0, 2π).

The solutions of the given equation are x = π/3, x = π/2, x = 2π/3, x = 4π/3, x = 3π/2 and x = 5π/3 in the interval [0, 2π).

Answer 2

• 15). The trigonometric equation is 3sec² (x) tan x = 4tan(x).

tan x[ 3sec² (x) - 4] = 0

⇒tan x = 0 and 3sec² (x) - 4 = 0.

• tan x = 0.

tan x = tan 0.

The genaral solution of tan(θ) = tan(α) is θ = nπ + α, where n is an integer.

⇒ x = nπ + 0

x = nπ.

If n = 0, x = (0)π = 0,

If n = 1, x = (1)π = π,

If n = 2, x = (2)π= 2π.

Therefore, the solutions of the given equation are x = 0 and x = π in the interval [0, 2π).

• 3sec² (x) - 4 = 0

3sec² (x) = 4

Using reciprocal identity : sec² x = 1/cos² x .

3(1/cos² x) = 4

3 = 4cos² x

cos² x = 3/4

cos x = ±√3/2.

⇒ cos x = √3/2  and  cos x = - √3/2.

• cos (x) = √3/2.

cos (x) = cos(π/6)

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

⇒ x = 2nπ ± π/6

If n = 0, x = 2(0)π + (π/6) and x = 2(0)π - (π/6) = π/6 and - π/6.

If n = 1, x = 2(1)π + (π/6) and x = 2(1)π - (π/6) = 2π + π/6 and 2π - π/6 = 13π/6 and 11π/6,

Therefore, the solutions of the given equation are x = π/6 and x = 11π/6 in the interval [0, 2π).

• cos (x) = - √3/2.

cos (x) = cos(5π/6)

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

⇒ x = 2nπ ± 5π/6

If n = 0, x = 2(0)π + (5π/6) and x = 2(0)π - (5π/6) = 5π/6 and - 5π/6.

If n = 1, x = 2(1)π + (5π/6) and x = 2(1)π - (5π/6) = 2π + 5π/6 and 2π - 5π/6 = 17π/6 and 7π/6,

Therefore, the solutions of the given equation are x = 5π/6 and x = 7π/6 in the interval [0, 2π).

The solutions of the given equation are x = 0, x = π/6, x = 5π/6, x = π, x = 7π/6 and x = 11π/6 in the interval [0, 2π).

Answer 3

• 16). The trigonometric equation is 4sin² (2x) = 1.

Divide each side by 4.

sin² (2x) = 1/4

⇒ sin (2x) = ± 1/2.

sin (2x) = 1/2 and sin (2x) = - 1/2.

• sin (2x) = 1/2.

sin(2x) = sin(π/6)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

 2x = nπ + (- 1)ⁿπ/6

⇒ x = [nπ + (- 1)ⁿπ/6]/2

If n = 0, x = [(0)π + (- 1)⁰π/6]/2 = (π/6)/2 = π/12,

If n = 1, x = [(1)π + (- 1)¹π/6]/2 = [π - π/6]/2 = (5π/6)/2 = 5π/12,

If n = 2, x = [(2)π + (- 1)²π/6]/2 = [2π + π/6]/2 = (13π/6)/2 = 13π/12,

If n = 3, x = [(3)π + (- 1)³π/6]/2 = [3π - π/6]/2 = (17π/6)/2 = 17π/12.

Therefore, the solutions of the given equation are x = π/12, x = 5π/12, x = 13π/12 and x = 17π/12 in the interval [0, 2π).

• sin (x) = - 1/2.

sin(2x) = sin(- π/6)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

2x = nπ + (- 1)ⁿ(- π/6)

⇒ x = [nπ + (- 1)ⁿ(- π/6)]/2.

If n = 0, x = [(0)π + (- 1)⁰(- π/6)]/2 = (- π/6)/2 = - π/12,

If n = 1, x = [(1)π + (- 1)¹(- π/6)]/2 = (π + π/6)/2 = (7π/6)/2 = 7π/12,

If n = 2, x = [(2)π + (- 1)²(- π/6)]/2 = (2π - π/6)/2 = (11π/6)/2 = 11π/12,

If n = 3, x = [(3)π + (- 1)³(- π/6)]/2 = (3π + π/6)/2 = (19π/6)/2 = 19π/12,

If n = 4, x = [(4)π + (- 1)⁴(- π/6)]/2 = (4π - π/6)/2 = (23π/6)/2 = 23π/12.

Therefore, the solutions of the given equation are x = 7π/12, x = 11π/12, x = 19π/12 and x = 23π/12 in the interval [0, 2π).

The solutions of the given equation are x = π/12, x = 5π/12, x = 7π/12, x = 11π/12, x = 13π/12, x = 17π/12, x = 19π/12 and x = 23π/12 in the interval [0, 2π).

Answer 4

• 17). The trigonometric equation is 8 - 12sin² (x) = 4 cos² (x).

Pythagorean identity : sin² x + cos² x = 1.

8 - 12(1 - cos² x) = 4 cos² x

8 - 12 + 12cos² x - 4 cos² x  = 0

8cos² x  - 4 = 0

4(2cos² x  - 1) = 0

⇒ 2cos² x  - 1 = 0

cos x = ± 1/√2.

• cos (x) = 1/√2..

cos (x) = cos(π/4)

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

⇒x = 2nπ ± (π/4)

If n = 0, x = 2(0)π + (π/4) and x = 2(0)π - (π/4) = π/4 and - π/4.

If n = 1, x = 2(1)π + (π/4) and x = 2(1)π - (π/4) = 2π + π/4 and 2π - π/4 = 9π/4 and 7π/4,

Therefore, the solutions of the given equation are x = π/4 and x = 7π/4 in the interval [0, 2π).

• cos (x) = - 1/√2.

cos (x) = cos(3π/4)

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

⇒x = 2nπ ± (3π/4)

If n = 0, x = 2(0)π + (3π/4) and x = 2(0)π - (3π/4) = 3π/4 and - 3π/4.

If n = 1, x = 2(1)π + (3π/4) and x = 2(1)π - (3π/4) = 2π + 3π/4 and 2π - 3π/4 = 12π/4 and 5π/4,

Therefore, the solutions of the given equation are x = 3π/4 and x = 5π/4 in the interval [0, 2π).

The solutions of the given equation are x = π/4, x = 3π/4, x = 5π/4 and x = 7π/4 in the interval [0, 2π).

• 18). The trigonometric equation is 2cos² x - 7cos x - 4 = 0.

Let, cos x = t.

Then, the trigonometric equation is 2t² - 7t - 4 = 0.

By factor by grouping.

2t² - 8t + t - 4 = 0

2t(t - 4) + 1(t - 4) = 0

(t - 4)(2t + 1) = 0

t - 4 = 0  and  2t + 1 = 0

t = 4  and  t = - 1/2

Put, t = cos x.

cos x = 4  and  cos x = - 1/2

• cos x = 4

cos x = cos(cos^{- 1}(4)).

⇒ x = cos^{- 1}(4).

It does not exists, since the range of the cosine function is [- 1, 1].

• cos (x) = - 1/2.

cos (x) = cos(2π/3)

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

⇒x = 2nπ ± (2π/3)

If n = 0, x = 2(0)π + (2π/3) and x = 2(0)π - (2π/3) = 2π/3 and - 2π/3,

If n = 1, x = 2(1)π + (2π/3) and x = 2(1)π - (2π/3) = 2π + 2π/3 and 2π - 2π/3 = 8π/3 and 4π/3.

⇒ x = 2π/3 and x = 4π/3 in the interval [0, 2π).

The solutions of the given equation are x = 2π/3 and x = 4π/3 in the interval [0, 2π).

Answer 5

• 19). The trigonometric equation is tan (2x) + sin x = 0.

[(sin 2x)/(cos 2x)] + sin x = 0

sin (2x) + sin x(cos 2x) = 0

Double angle formulas : sin(2x) = 2 sin x cos x,

                                           cos(2x) = 2cos² x - 1.

2 sin x cos x + sin x(2cos² x - 1) = 0

sin x[2cos² x + 2 cos x - 1] = 0

sin x = 0  and  2cos² x + 2 cos x - 1 = 0

• sin (x) = 0.

sin(x) = sin(0)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

⇒ x = nπ + (- 1)ⁿ(0)

If n = 0, x = (0)π + (- 1)⁰0 = 0.

If n = 1, x = (1)π + (- 1)¹0   π - 0  = π.

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

• 2cos² x + 2cos x - 1 = 0

Let, cos x = t.

Then, the equation is 2t² + 2t - 1 = 0.

2t² + 2t - 1 = 0, is quadratic, use quadratic formula to find the roots of the related quadratic equation.

Solution t = [- b ± √ (b² - 4ac)]/2a.

Compare the equation with standard form of quadratic equation : ax² + bx + c = 0.

a = 2, b = 2, and c = - 1.

Solution t = [- 2 ± √ (2² - 4*2*(- 1))]/2*2

t = [- 2 ± √(4 + 8)]/4

t = [- 2 ± √12]/4

t = [- 2 ± 2√3]/4

t = [- 1 ± √3]/2

t = [- 1 - √3]/2  and  y = [- 1 + √3]/2

Put, t = cos x.

cos x = [- 1 - √3]/2  and  cos x =[- 1 + √3]/2.

cos x = cos(cos^{- 1}([- 1 - √3]/2) and  cos x = cos(cos^{- 1}([- 1 + √3]/2)

⇒ x = cos^{- 1}([- 1 - √3]/2) and  x = cos^{- 1}([- 1 + √3]/2).

x = cos^{- 1}([- 1.366) and x = cos^{- 1}(0.3660).

x = cos^{- 1}([- 1.366) does not exists, since the range of the cosine function is [- 1, 1].

The solutions is x = cos^{- 1}([- 1 +√3]/2) in the interval [0, 2π).

Therefore, the solutions of the given equation are x = 0, x = π and  x = cos^{- 1}([- 1 +√3]/2) in the interval [0, 2π).

Answer 6

• 20). The trigonometric equation is sec (4x) + 2sin (4x) = 0.

Using reciprocal identity : sec x = 1/cos x.

[1/cos (4x)] + 2sin (4x) = 0

1 + 2 sin (4x) cos (4x) = 0

Use the formula : 2 sin nx cos nx = sin (2nx).

1 + sin(8x) = 0

sin(8x) = - 1

sin(8x) = sin(- π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

8x = nπ + (- 1)ⁿ(- π/2)

⇒ x = [nπ + (- 1)ⁿ(- π/2)]/8

If n = 0, x = [(0)π + (- 1)⁰(- π/2)]/8 = (- π/2)/8 = - π/16,

If n = 1, x = [(1)π + (- 1)¹(- π/2)]/8 = [π + π/2]/8 = (3π/2)/8 = 3π/16,

If n = 2, x = [(2)π + (- 1)²(- π/2)]/8 = [2π - π/2]/8 = (3π/2)/8 = 3π/16,

If n = 3, x = [(3)π + (- 1)³(- π/2)]/8 = [3π + π/2]/8 = (7π/2)/8 = 7π/16,

If n = 4, x = [(4)π + (- 1)⁴(- π/2)]/8 = [4π - π/2]/8 = (7π/2)/8 = 7π/16,

If n = 5, x = [(5)π + (- 1)⁵(- π/2)]/8 = [4π + π/2]/8 = (9π/2)/8 = 9π/16,

If n = 6, x = [(6)π + (- 1)⁶(- π/2)]/8 = [6π - π/2]/8 = (11π/2)/8 = 11π/16,

If n = 7, x = [(7)π + (- 1)⁷(- π/2)]/8 = [7π + π/2]/8 = (15π/2)/8 = 15π/16,

If n = 8, x = [(8)π + (- 1)⁸(- π/2)]/8 = [8π - π/2]/8 = (15π/2)/8 = 15π/16,

If n = 9, x = [(9)π + (- 1)⁹(- π/2)]/8 = [9π + π/2]/8 = (19π/2)/8 = 19π/16,

If n = 10, x = [(10)π + (- 1)¹⁰(- π/2)]/8 = [10π - π/2]/8 = (19π/2)/8 = 19π/16,

If n = 11, x = [(11)π + (- 1)¹¹(- π/2)]/8 = [11π + π/2]/8 = (23π/2)/8 = 23π/16,

If n = 12, x = [(12)π + (- 1)¹²(- π/2)]/8 = [12π - π/2]/8 = (23π/2)/8 = 23π/16,

If n = 13, x = [(13)π + (- 1)¹³(- π/2)]/8 = [13π + π/2]/8 = (27π/2)/8 = 27π/16,

If n = 14, x = [(14)π + (- 1)¹⁴(- π/2)]/8 = [14π - π/2]/8 = (27π/2)/8 = 27π/16,

If n = 15, x = [(15)π + (- 1)¹⁵(- π/2)]/8 = [15π + π/2]/8 = (31π/2)/8 = 31π/16.

Therefore, the solutions of the given equation are x = 3π/16, x = 7π/16, x = 9π/16, x = 11π/16, x = 15π/16, x = 19π/16, x = 23π/16, x = 27π/16 and x = 31π/16 in the interval [0, 2π).

Answer 7

• 21). The trigonometric equation is sin (2x) - sin (x) = 2cos x - 1.

Double angle formula : sin(2x) = 2 sin x cos x.

2 sin x cos x - sin (x) = 2cos x - 1

sin x[2 cos x - 1] = [2cos x - 1]

Subtract [2 cos x - 1] from each side.

sin x[2 cos x - 1] - [2cos x - 1] = 0

[2 cos x - 1](sin x - 1) = 0

2 cos x - 1 = 0  and  sin x - 1 = 0

• 2 cos x - 1 = 0.

cos x = 1/2

cos (x) = cos(π/3)

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

⇒x = 2nπ ± (π/3)

If n = 0, x = 2(0)π + (π/3) and x = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, x = 2(1)π + (π/3) and x = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

⇒ x = π/3 and x = 5π/3 in the interval [0, 2π).

• sin x - 1 = 0

sin x = 1

sin(x) = sin(π/2)

The genaral solution of sin(θ) = sin(α) is θ = nπ + (- 1)ⁿα, where n is an integer.

⇒ x = nπ + (- 1)ⁿ(π/2)

If n = 0, x = (0)π + (- 1)⁰(π/2) = π/2,

If n = 1, x = (1)π + (- 1)¹(π/2) = π - π/2 = π/2,

If n = 2, x = (2)π + (- 1)²(π/2) = 2π + π/2 = 5π/2,

⇒ x = π/2 in the interval [0, 2π).

The solutions of the given equation are x = π/3, x = π/2,  and x = 5π/3 in the interval [0, 2π).

Answer 8

• 22). The trigonometri equation is sin^{- 1} x - cos^{- 1} x = π/6.

Inverse trigonometric function : sin^{- 1} x + cos^{- 1} x = π/2.

(π/2 - cos^{- 1} x) - cos^{- 1} x = π/6

- 2cos^{- 1} x = π/6 - π/2

- 2cos^{- 1} x = (π - 3π)/6

- 2cos^{- 1} x = - 2π/6

- 2cos^{- 1} x = - π/3

cos^{- 1} x = π/6

⇒ x = cos(π/6) = √3/2.

The solution of the given equation is x = √3/2.

• 24). The trigonometric equation is cos^{- 1} x + 2sin^{- 1} (√3/2) = π/3.

cos^{- 1} x + 2sin^{- 1} (sin(π/3)) = π/3

sin(sin^-1(x)) = x   for every x in the interval [-1, 1].

sin^-1(sin(x)) = x   for every x in the interval [ - π/2, π/2].

cos^{- 1} x + (2 * π/3) = π/3

cos^{- 1} x + 2π/3 = π/3

cos^{- 1} x = π/3 - 2π/3

cos^{- 1} x = (π - 2π)/3 = - π/3

⇒ x = cos(- π/3) = cos(π - π/3) = - 1/2.

The solution of the given equation is x = - 1/2.

• 25). The trigonometri equation is  cos^{- 1} x = sin^{- 1} x.

Inverse trigonometric function : sin^{- 1} x + cos^{- 1} x = π/2.

π/2 - sin^{- 1} x = sin^{- 1} x

π/2 = 2sin^{- 1} x

sin^{- 1} x = π/4

⇒ x = sin(π/4) = 1/√2.

The solution of the given equation is x = 1/√2.