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alexira [117]
1 year ago
14

Find the solution(s) to the equation sine (startfraction x over 2 endfraction) cosine x minus 1 = 0 on the interval 0 ≤ x 2π.

check all that apply. 0 startfraction pi over 3 endfraction startfraction pi over 6 endfraction startfraction 3 pi over 2 endfraction startfraction 5 pi over 3 endfraction
Mathematics
1 answer:
Katyanochek1 [597]1 year ago
3 0

All apply values to the equation are:

0 ⇒ A

π/3 ⇒ B

5π/3 ⇒ E

What is an equation?

A formula known as an equation shows the equality of two expressions by joining them with the equals sign =.

Main Body:

∵ 0 ≤ x ≤ 2π is the domain for angle x

∴ 0 ≤x/2  ≤ π is the domain of angle

∵ sin(x/2) + cos(x) - 1 = 0

→ To solve the equation we should use the rule of cosine double angle

∵ cos(x) = 1 - 2 sin²(x/2)

→ Substitute it in the equation above

∴ sin(x/2) + (1 - 2 sin²(x/2) = 0

∴ sin(x/2) + 1 - 2 sin²(x/2) = 0

→ Add the like terms

∴ sin(x/2) + (1 - 1) - 2 sin²(x/2) = 0

∴ sin(x/2) - 2 sin²(x/2) = 0

→ Take sin(x/2) as a common factor

∴ sin(x/2) [1 - 2sin(x/2)] = 0

→ Equate each factor by 0

∵ sin(x/2) = 0

→ The value of sine equal zero on the x-axis

∴ x/2 = 0, π, 2π

∵ The domain of x/2  is 0 ≤ (x/2) ≤ π

∴  x/2 = 0 and π ⇒ 2π refused because ∉ the domain

→ Multiply both sides by 2 to find x

∴ x = 0 and 2π

∵ 1 - 2sin(x/2) = 0

→ Subtract 1 from both sides

∴ - 2sin(x/2) = -1

→ Divide both sides by -2

∴ sin(x/2) = 1/2

→ The sine is positive in the 1st and 2nd quadrants

∴ (x/2) lies on the 1st OR 2nd quadrants

∵ (x/2) = sin⁻¹(1/2)

∴ (x/2) =  π/6 =  ⇒ 1st quadrant

→ Multiply both sides by 2

∴ x = π/3

∵ (x/2) = π -π/6  = 5π/6 ⇒ 2nd quadrant

→ Multiply both sides by 2

∴ x = 5π/3

∴ The values of x are 0,π/3 , 5π/3 , 2π

Hence  apply values are:

0 ⇒ A

π/3⇒ B

5π/3⇒ E

To learn more about equation ,visit:

brainly.com/question/22688504

#SPJ4

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