JEE Class main Answered

5 cos2θ - 2 sin2θ = 2 ................... (1)

5 ( 2cos²θ - 1) - 2 ( 2 sinθ cosθ ) = 2

10 cos²θ - 4 sinθ cosθ - 7 = 0

10 cos²θ - 7 =  4 ( 1-cos²θ )^1/2 cosθ

By squaring both sides , we get

100 cos⁴θ +49 - 140 cos²θ =  16 ( 1-cos²θ ) cos²θ

116 cos⁴θ  - 156 cos²θ + 49 =  0

116 cos⁴θ  - 58 cos²θ - 98 cos²θ + 49 =  0

58 cos²θ( 2 cos²θ- 1 ) - 49( 2 cos²θ- 1 ) = 0

( 2 cos²θ- 1 ) ( 58 cos²θ - 49 ) = 0

Hence cos²θ = 1/2  or 49/58  

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If cos²θ = 1/2 , then cosθ = ±1/√2

 θ = ± [ (π/4)+2nπ ] or ± [ (7/4)π+2nπ ]

If we substitute θ in eqn.(1) and verify eqn.(1) , we get

θ = - [ (π/4)+2nπ ] or + [ (7/4)π+2nπ ]

( Other values of θ are not matching and can be discarded )

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If cos²θ = 49/58 , then cosθ = ± 0.9192

θ = ±0.4049 or ± (2π-0.4049)

If we substitute θ in eqn.(1) and verify eqn.(1) , we get

θ = +0.4049 or -(2π-0.4049)

( Other values of θ are not matching and can be discarded )

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