Question

if 4cos²θ - 8cosθ + 3=0 , then θ is equal to​

Answer 1

Answer:

I am representing theta as A because I don't have theta

4cos²A - 8cosA + 3 = 0

or, 4cos²A - (6+2)cosA + 3 = 0

or, 4cos²A - 6cosA -2cosA + 3 = 0

or, 2cosA(2cosA - 3)-1(2cosA-3) = 0

or, (2cosA - 1)(2cosA-3) = 0

either 2cosA - 1 = 0

then cosA= 1/2

or 2cosA - 3 = 0

then cos A = 3/2

But value of cos A can't be 3/2

because cos 30° = √3/2

So cos A = 1/2 = 60°

Value of A or theta is 60°