Find approximate solution for the equation in the interval -π tan x= 2 csc x

The angle in the 1st quadrant is 1.144 and in the 4th quadrant is -1.144

∴ The answer is (a)

Step-by-step explanation:

* The domain of the function is -π < x < π

- Then the angle is in the 1st or 4th quadrant

∵ tan(x) = 2 csc(x)

∵ tan(x) = sin(x)/cos(x)

∵ csc(x) = 1/sin(x)

∴ sin(x)/cos(x) = 2(1/sin(x)) = 2/sin(x) ⇒ using cross multiplication

∴ sin²(x) = 2cos(x)

∵ sin²(x) = 1 - cos²(x) ⇒ substitute it in the last step

∴ 1 - cos²(x) = 2cos(x) ⇒ arrange the terms in one side

∴ cos²(x) + 2cos(x) - 1 = 0

* Lets factorize it using the formula

∵ a = 1 , b = 2 , c = -1

∵ x = [-b ± √(b² - 4ac)]/2(a) ⇒ formula of quadratic equation

∵ b² - 4ac = 2² - 4(1)(-1) = 4 - -4 = 4 + 4 = 8

∵ √8 = 2√2

∴ cos(x) = [-2 ± 2√2]/2(1) = [-2 ± 2√2]/2 ⇒ ÷ 2 up and down

∴ cos(x) = -1 ± √2

* cos(x) = -1 + √2 ⇒ positive value and cos(x) = -1 - √2 ⇒ negative value

∵ x lies on 1st or 4th quadrant

∴ cos(x) must be positive according to the ASTC rule

∴ We will rejected the negative value

* Now lets find the values of angle x

∵ cos(x) = -1 + √2

∴ x = cos^-1(-1 + √2) = 1.1437 ≅ 1.144 ⇒ approximated to the nearest

  3 decimal place

* The angle in the 1st quadrant is 1.144 and in the

  4th quadrant is-1.144

∴ The answer is (a)

h is the x-coordinate of the vertex

step-by-step explanation:

we have

in this function

the point represent the vertex of the function

so

h is the x-coordinate of the vertex

k is the y-coordinate of the vertex