Question

6cos^2 theta + 5cos theta=1
Solve for theta

Answer 1

Substitute :

cos ( theta ) = u

6 cos ^ 2 ( theta ) + 5 cos ( theta ) = 1

6 u ^ 2 + 5 u = 1

6 u ^ 2 + 5 u - 1 = 0

The exact solutions are :

u = 1 / 6

cos ( theta ) = 1 / 6

theta = cos ^ - 1 ( 1 / 6 )

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cos ^ - 1

is the inverse cosine function

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and


u = - 1

cos ( theta ) = - 1

theta = pi [ radians ] = 180 °

The cosine function have periode 2 pi n

where nis an integer.


So the solutions are :

theta = 2 pi n + cos ^ - 1 ( 1 / 6 )

theta = 2 pi n - cos ^ - 1 ( 1 / 6 )


and


theta = 2 pi n + pi

theta = 2 pi n - pi



P.S.

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6 u ^ 2 + 5 u - 1 = 0

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6 u ^ 2 + 5 u - 1 = 0

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