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.
Since there is no equation, there is nothing that can be solved.
Zero. Anything minus itself is zero.
It is 1.
It is a mathematical expression.
If sin2(theta) = 0, then theta is N pi, N being any integer
There is no easy simplification.
Since there is no equation, there is nothing that can be solved.
cos2(theta) = 1 so cos(theta) = ±1 cos(theta) = -1 => theta = pi cos(theta) = 1 => theta = 0
Zero. Anything minus itself is zero.
It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.
cosine (90- theta) = sine (theta)
-2(cot2theta)
Tan^2
If r-squared = theta then r = ±sqrt(theta)
-Sin^(2)(Theta) + Cos^(2)Theta => Cos^(2)Theta - Sin^(2)Theta Factor (Cos(Theta) - Sin(Theta))( Cos(Theta) + Sin(Theta)) #Is the Pythagorean factors . Or -Sin^(2)Theta = -(1 - Cos^(2)Theta) = Cos(2)Theta - 1 Substitute Cos^(2)Thetqa - 1 + Cos^(2) Theta = 2Cos^(2)Theta - 1
It is 1.
It is a mathematical expression.