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Assume the first term is cos-squared(theta) rather than cos(2*theta).

6cos2(t) + 5cos(t) - 4 = 0

[6cos2(t) + 8cos(t) - 3 cos(t) - 4] = 0

[3cos(t) + 4]*[2cos(t) - 1] = 0

which gives

3cos(t) = -4 so that cos(t) = -4/3

or

2cos(t) = 1 so that cos(t) = 1/2

The first of these is clearly not a possible solution, whereas the second can hve one or more solutions, depending on the domain - which is not specified in the question.

If the initial assumption is incorrect and the first term WAS 6cos(2*theta) then using the formula for double angles:

cos(2(t) = cos2(t) - sin2(t) = cos2(t) - [1 - cos2(t)] = 2cos2(t) - 1

and so, the equation becomes

6*[2cos2(t) - 1] + 5cos(t) - 4 = 0

or 12cos2(t) + 5cos(t) - 10 = 0

Solve this quadratic equation for cos(t) and then find the values of t (or theta) within the domain specified.

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13y ago

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