Solve each equation for exact solutions over the interval [0,2pi]

Question

1. 2cos^2 x-7 cos x= 3 

2. cos x/2 - cos x= 1 

3. 4 sin ^2 x+ 4 cos x-5= 0 

4. 2 cos^2 2x+ 3 cos 2x +1=0 

5. 3 sec^2x tan x = 4tan x 

6. 4sin^2x=1 

7. 8-12 sin^2 x= 4 cos^2 x

 

Answer 1

• 1).

The trigonometric equation is 2cos² x - 7cos x = 3.

Let, cos x = y.

Then, the trigonometric equation is 2y² - 7y - 3 = 0.

2y² - 7y - 3 = 0, is quadratic, use quadratic formula to find the roots of the related quadratic equation.

Solution x = [- b ± √ (b² - 4ac)]/2a.

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

a = 2, b = - 7, and c = - 3.

Solution y = [- (- 7) ± √ ((- 7)² - 4*2*(- 3))]/2*2

y = [7 ± √(49 + 24)]/4

y = [7 ± √73]/4.

y = [7 + √73]/4  and  y = [7 - √73]/4

Put, y = cos x.

cos x = [7 + √73]/4  and  cos x = [7 - √73]/4

cos x = cos(cos^{- 1}([7 + √73]/4)) and  cos x = cos(cos^{- 1}([7 - √73]/4))

⇒ x = cos^{- 1}([7 + √73]/4) and  x = cos^{- 1}([7 - √73]/4).

The solution of the given equation is x = cos^{- 1}([7 + √73]/4) and  x = cos^{- 1}([7 - √73]/4) in the interval [0, 2π).

• 2).

The trigonometric equation is cos (x/2) - cos x = 1.

Half - angle formula : cos(x/2) = ±√[(1 + cos x)/2].

±√[(1 + cos x)/2] - cos x = 1

±√[(1 + cos x)/2] = 1 + cos x

Squaring on each side.

( ±√[(1 + cos x)/2] )² = (1 + cos x)²

(1 + cos x)/2 = 1 + 2cos x + cos² x

1 + cos x = 2 + 4cos x + 2cos² x

2 + 4cos x - cos x + 2cos² x - 1 = 0

2cos² x  + 3cos x + 1= 0

2cos² x  + 2cos x + cos x + 1= 0

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

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

⇒ cos x + 1 = 0  and  2cos x + 1 = 0

cos x = - 1  and  cos x = - 1/2

cos x = cos π  and  cos x = cos(2π/3) in the interval [0, 2π).

⇒ x = π and x = 2π/3.

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

• 3). 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,

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

Answer 2

• 4). 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 3

• 5). 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 4

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

Divide each side by 4.

sin² (x) = 1/4

⇒ sin x = ± 1/2.

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

• sin (x) = 1/2.

sin(x) = sin(π/6)

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

⇒ x = nπ + (- 1)ⁿπ/6

If n = 0, x = (0)π + (- 1)⁰π/6 = π/6.

If n = 1, x = (1)π + (- 1)¹π/6   π - π/6  = 5π/6.

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

• sin (x) = - 1/2.

sin(x) = sin(- π/6)

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

⇒ x = nπ + (- 1)ⁿ(- π/6)

If n = 0, x = (0)π + (- 1)⁰(- π/6) = - π/6.

If n = 1, x = (1)π + (- 1)¹(- π/6) = π + π/6  = 7π/6.

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

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

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

Answer 5

• 7). 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).