Question

This is hard, very very hard!

Answer 1

this is not hard ,

5cos2∅ = 15cos∅ -10

use cos2∅ = 2cos²∅ -1
5(2cos²∅ -1) = 15cos∅ -10
10cos²∅ -5 = 15cos∅ -10
10cos²∅ -15cos∅ +5 =0
5(2cos²∅ -3cos∅ +1) =0
2cos²∅ -2cos∅ -cos∅ +1 =0
(cos∅ -1)( 2cos∅ -1) =0
cos∅ = 1, 1/2

now ,given ∅ belongs to [0, 2π)

cos∅ = 1 when, ∅ = 0
cos∅ = 1/2 when , ∅ = π/3 , 5π/3

hence , ∅ = 0, π/3, 5π/3

Answer 2

Answer:

this is not hard ,

5cos2∅ = 15cos∅ -10

use cos2∅ = 2cos²∅ -1

5(2cos²∅ -1) = 15cos∅ -10

10cos²∅ -5 = 15cos∅ -10

10cos²∅ -15cos∅ +5 =0

5(2cos²∅ -3cos∅ +1) =0

2cos²∅ -2cos∅ -cos∅ +1 =0

(cos∅ -1)( 2cos∅ -1) =0

cos∅ = 1, 1/2

now ,given ∅ belongs to [0, 2π)

cos∅ = 1 when, ∅ = 0

cos∅ = 1/2 when , ∅ = π/3 , 5π/3

hence , ∅ = 0, π/3, 5π/3

Step-by-step explanation: