Question

1 + 2 cos 3x cos X - cos 2x = 0.​

Answer 1

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Answer 2

Answer:

1+2×(4cos^3(x) -3cosx)cosx - cos2x = 0

=> 1+8cos^4(x) - 6cos^2(x) - (2cos^2x - 1) =0

=> 1+8cos^4(x) - 8cos^2(x) +1= 0

=> 8cos^4(x) - 8cos^2(x) +2 =0

=> 4cos^4(x) - 4 cos^2(x) +1= 0

=> 4cos^2(x)× {cos^2(x) -1 }+ 1 = 0

=> 4cos^2(x)× {1 - cos^2(x) }- 1 =0

=> 4 cos^2(x).sin^2(x) -1 = 0

=> (2 cosx. sinx)^2 - 1=0

=> sin^2(2x) = 1

=> sin2x = +- 1 , but sin90° & sin 180° is always positive so sin2x = 1 here

=>sin2x = sin 90 or sin 180

=> 2x = 90 , 2x = 180

=> x= 45 or x= 90