trigonometry math

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Find the exact solutions of the equation cos 2a=square root(1-sin^2 a), 0<a<180.

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We have to determine the solutions for cos 2a = sqrt( 1 - sin a)^2, lying in the range 0<a<180

cos 2a = sqrt( 1 - (sin a)^2)

=> cos 2a = sqrt ( (cos a)^2)

=> cos 2a = cos a

=> 2*(cos a)^2 - 1 = cos a

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We have to determine the solutions for cos 2a = sqrt( 1 - sin a)^2, lying in the range 0<a<180

cos 2a = sqrt( 1 - (sin a)^2)

=> cos 2a = sqrt ( (cos a)^2)

=> cos 2a = cos a

=> 2*(cos a)^2 - 1 = cos a

=> 2*(cos a)^2 - cos a - 1 = 0

=> 2*(cos a)^2 - 2cos a + cos a - 1 = 0

=> 2*cos a ( cos a + 1) + 1(cos a + 1) = 0

=> (2cos a + 1)( cos a + 1) = 0

2*cos a + 1 = 0

=> cos a = -1/2

=> a = arc cos (-1/2)

=> 120 degrees

cos a = 1

=> a = arc cos (1)

=> a = 0, but it does not lie in the interval.

The required value of a is 120 degrees.

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