Question

Solve the equation: cos 2θ + cos 8θ = cos 5θ

Answer 1

ANSWER:⇒ The Answer is θ = π/10 , θ = 3π/10 , θ = π/9

& θ = 5π/9

Given that

cos 2θ + cos 8θ = cos 5θ

cos 8θ + cos 2θ  = cos 5θ

2[(cos 8θ + cos 2θ)/2][(cos 8θ - cos 2θ)/2]  = cos 5θ

2[(cos 10θ)/2][(cos 6θ)/2]  = cos 5θ

2[cos 5θ][cos 3θ]  = cos 5θ

2[cos 5θ][cos 3θ]  - cos 5θ = 0

cos 5θ [ 2 cos 3θ - 1 ]  = 0

cos 5θ = 0 AND  cos 3θ - 1  = 0

For cos 5θ = 0 : -

cos 5θ = 0

5θ  = π/2 , 5θ = 3π/2

θ  = π/10 , θ = 3π/10

For cos 3θ - 1  = 0 : -

2 cos 3θ  = 1

cos 3θ  = 1/2

The Reference Angle is π/3

Cos is +ve in Quadrant I & IV

For Quad I : -

3θ = π/3

θ = π/9

For Quad IV : -

3θ = 2π - π/3 = 5π/3

θ = 5π/9

⇒ The Answer is θ = π/9 & θ = 5π/9

Answer 2

Solution :

Here I am using A instead of theta.

Given cos2A + cos8A = cos 5A


=> cos2A+cos8A-cos5A= 0

=>2cos[(2A+8A)/2]cos[(2A-8A)/2]-cos5A=0

=>2cos5Acos3A-cos5A=0

=>cos5A(2cos3A-1)=0

=>cos5A=0 or 2cos3A-1=0

i ) If cos5A=0 then

5A= (2n+1)π/2

=> A = (2n+1)π/10 , n €Z

ii ) If 2cos3A-1 = 0

then

cos3A = 1/2 = cosπ/3

=> 3A = 2nπ±π/3

=> A = [(2nπ)/3]± π/9 , n €Z

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