Question

If 32 tan^8 theta = 2 cos^2 alpha - 3 cos alpha and 3 cos 2theta=1 then the sum of values of alpha is

Answer 1

Answer:

Step-by-step explanation:

Given If 32 tan^8 theta = 2 cos^2 alpha - 3 cos alpha and 3 cos 2 theta = 1 then the sum of values of alpha is

Given 3 cos 2 theta = 1

So cos 2 theta = 1/3  

  We know that cos 2 theta

= 1 – tan^2 theta / 1 + tan^2 theta = 1/3

3– 3 tan^2 theta = 1 + tan^2 theta

2 = 4 tan^2 theta

Tan^2 theta = 1/2

Now 32 (tan^2 theta)^4 = 2 cos^2 α – 3 cos α

32 (1 / 2)^4 = 2 cos^2 α – 3 cos α

2 cos^2 α – 3 cos α – 2 = 0

let cos α = x

so we get

x^2 – 3 x – 2 = 0

2x^2 – 4 x + x – 2 = 0

2 x(x – 2) + 1(x – 2) = 0

(2x + 1)(x – 2) = 0

x = 2, x = - 1/2

    cos ∝ = 2, - 1/2

Taking cos ∝ = - 1/2 we get

So α = 2nπ + 2π / 3

Answer 2

Answer:

$\alpha = 2n\pi + 2\pi \div 3$