Maths Equation Question 2x(sq)-x-6=0?

Let a = arccos x, then x = cos a, and 2x = 2 cos a , and sin a = sqrt(1 - x^2) If be = arcsin(2x), then 2x = sin be = 2 cos a, and cos be = (1 - 4 x^2 )^(1/2) Taking the sin() of both sides sin( a + b) = 1/2 sin a cos be + cos a sin be = 1/2 (1 - x^2)^(1/2) (1 - 4x^2)^(1/2) + 2x^2 = 1/2 Hence (1 - x^2) (1 - 4x^2) = 1/4 - 2 x^2 + 4x^4 1 - 5 x^2 + 4x^4 = 1/4 - 2 x^2 + 4x^4 3 x^2 = 3/4 From which x = 1/2, or x = -1/2 Plugging in the in the original equation we get.

( pi / 6 ) - arccos x = arcsin (2x ) sin ( pi / 6 ) - arccos x = 2x sin ( pi / 6 ) cos ( arccos x ) - cos ( pi / 6 ) sin ( arccos x ) = 2x ( 1 / 2 )x - sqrt ( 3 ) / 2 sqrt ( 1 - x^2 ) = 2x sqrt ( 3 ) / 2 sqrt ( 1 - x^2 ) = ( - 3 / 2 )x sqrt ( 1 - x^2 ) = - sqrt ( 3 ) x 1 - x^2 = 3x^2 1 = 4x^2 x^2 = 1 / 2 x = + - sqrt ( 2 ) / 2.

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