Question

8. Find all solutions for the following trig equation. 4cos? - 9cose + 2 = 0

Answer 1

Solution:

The values provided in the questions are 4cos²(θ) - 9cos(θ) + 2=0

Let  

4cos²(θ) - 9cos(θ) + 2=0 ------------------------------->1

Assume

cos(θ) = u

Substitute cos(θ) =u into equation 1

   4u² - 9u + 2 = 0 ----------------------------------------------->2

splitup the above equation 2 by the quadratic formula method

4u² - 8u - 1u +2 = 0

4u (u - 2 ) - 1 (u - 2) = 0

(4u - 1) ( u - 2) = 0

4u - 1 = 0 , u - 2 = 0

4u = 1 , u = 2

u = 1/4 , u = 2

put u = cos(θ) in above equation

cos(θ) = 1/4 , cos(θ) = 2 ==> θ = cos^-1(2)

we know that

θ = cos^-1(2) = none values

     cos(θ) = 1/4

θ = cos^-1(1/4)

we know that formula

cos(θ) = 1/4 = => θ = cos^-1(1/4) + 2πn ,  θ = θ = 2π - cos^-1(1/4) + 2πn

  θ = 1.318 + 2πn , θ = 2π - 1.318 +2πn    radian