Verified answer

tan2θ = sin2θ/cos2θ

also we know the following identities,

sin2θ = 2 sinθcosθ = 2 sinθ * sqrt(1 - sin²θ)

and cos2θ = 1 - 2 sin²θ

Let sinθ = x

So we get the original question as:

2 x sqrt(1-x²) ... + .... 2 x = 0

--------------------

1 - 2 x²

Simplifying this we get

2x sqrt(1-x²) + 2x (1-2x²) = 0

Basically we managed to get everything in the question in terms of one variable i.e. sin, and then solve for that

Whatever answer you get for x, you can find the sin inverse of that to get the value of theta.

All the best!

tan(2θ) + 2sin(θ) = 0

sin(2θ)/cos(2θ) = -2sin(θ) ................. note: tan(2θ) = sin(2θ)/cos(2θ)

sin(2θ)/cos(2θ) = -sin(2θ)/cos(θ) ....... note: sin(2θ) = 2 sin(θ)cos(θ)

cos(2θ) + cos(θ) = 0 ........................ discards θ=0, reclaim it later.

2cos²(θ) + cos(θ) - 1 = 0 .................. note: cos(2θ) = 2cos²(θ) - 1

(2cos(θ) - 1)(cos(θ) + 1) = 0

cos(θ) = 1/2 or cos(θ) = -1

θ = π/3, 5π/3, π ...

θ = 0 .................................................. discarded earlier and reclaimed here.

Answer: θ = 0, π(2n + 1/3), π(2n + 5/3), π(2n + 1) n ∈ ℤ

As to methodology, you can see there is more than one solution. flyingagent rearranged the equation in terms of sin(θ) such that sin(θ) = 0, ±√3/2. I rearranged in terms of cos(θ).

Take the sine of the two facets sin(arcsin(x) + arccos(x)) => sin(arcsin(x))cos(arccos(x)) + sin(arccos(x))cos(arcsin(x)) => x * x + sqrt(a million - x^2) * sqrt(a million - x^2) => x^2 + a million - x^2 => a million sin(arcsin(x) + arccos(x)) = a million arcsin(x) + arccos(x) = arcsin(a million) arcsin(x) + arccos(x) = pi/2