Question

Solve the equation and write general solution: 4 cos² θ + √3 = 2(√3 + 1) cos θ.

Answer 1

Answer:

general solution of given equation is θ= ±π/3+2πk ,±π/6+2πk,for any integer k.

Step-by-step explanation:

Solve the equation and write general solution: 4 cos² θ + √3 = 2(√3 + 1) cos θ

4 cos² θ - 2(√3 + 1) cos θ +√3 =0

4 cos² θ - 2√3 cos θ  -2 cos θ +√3 =0

2 cos θ(2 cos θ -√3)-1(2 cos θ -√3)=0

(2 cos θ -√3)(2 cos θ -1)=0

So

(2 cos θ -√3)=0

2 cos θ = √3

cos θ=√3/2

θ= cos⁻1( cos π/6)

θ= π/6

and

(2 cos θ -1)=0

cos θ=1/2

θ= cos⁻1( cos π/3)

θ= π/3

So general solution of given equation is θ= ±π/3+2πk ,±π/6+2πk,for any integer k.

Answer 2

Solution :

Here I am using A instead of theta.

4cos²A + √3 = 2(√3 + 1 ) cosA

=> 4cos²A+√3 = 2√3 cosA + 2cosA

=> 4cos²A + √3 -2√3cosA - 2cosA = 0

=> 4cos²A-2cosA -2√3cosA + √3 = 0

=> 2cosA(2cosA-1) -√3( 2cosA-1) = 0

=> (2cosA - 1 )( 2cosA - √3 ) = 0

2cosA - 1 = 0 or 2cosA - √3 = 0

cosA = 1/2 or cosA = √3/2

A = 2nπ ± ( π/3 ) or A = 2nπ±(π/6) ,n€Z

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=> 2cosA( 2cosA -1 )-√3( 2cosA + 2 )=0