There's one such solution. Try Θ = π / 6 in the equation, and you'll see that it works.
EDIT: If you don't want to take the lazy route, you could also find the general solution and pick a solution you like from it.
cos(2Θ) = sin(Θ)
cos²(Θ) - sin²(Θ) = sin(Θ)
1 - sin²(Θ) - sin²(Θ) = sin(Θ)
2sin²(Θ) + sin(Θ) - 1 = 0
This is a quadratic in sin(Θ). All we need to do is factorise:
(2sin(Θ) - 1)(sin(Θ) + 1) = 0
sin(Θ) = 1/2 or -1
That is, choosing any Θ such that sin(Θ) = 1/2 or -1 will be a solution to the original equation. Notice that Θ = π / 6 produces sin(Θ) = 1/2. We could also have chosen Θ = 3π / 2, or one of the infinitely many other solutions.
(E) Infinitely many This is because both equations are telling the same story and so in effect you only have one equation with 2 unknowns. The solutions fit on a straight line when graphed ... there being an infinite number of them
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Well, you know that:
cos(π / 2 - ϑ) = sin(ϑ)
so we could work from there. Let's assume that:
2Θ + Θ = π / 2
3Θ = π / 2
Θ = π / 6
There's one such solution. Try Θ = π / 6 in the equation, and you'll see that it works.
EDIT: If you don't want to take the lazy route, you could also find the general solution and pick a solution you like from it.
cos(2Θ) = sin(Θ)
cos²(Θ) - sin²(Θ) = sin(Θ)
1 - sin²(Θ) - sin²(Θ) = sin(Θ)
2sin²(Θ) + sin(Θ) - 1 = 0
This is a quadratic in sin(Θ). All we need to do is factorise:
(2sin(Θ) - 1)(sin(Θ) + 1) = 0
sin(Θ) = 1/2 or -1
That is, choosing any Θ such that sin(Θ) = 1/2 or -1 will be a solution to the original equation. Notice that Θ = π / 6 produces sin(Θ) = 1/2. We could also have chosen Θ = 3π / 2, or one of the infinitely many other solutions.
cos(2Θ)=sinΘ
Θ=0.5236 rad
Θ=30°
(E) Infinitely many This is because both equations are telling the same story and so in effect you only have one equation with 2 unknowns. The solutions fit on a straight line when graphed ... there being an infinite number of them
Θ = π/6
Θ = 5π/6
Θ = 3π/2
for 0 ≤ Θ ≤ 2π
And here's why
cos(2Θ)=sinΘ
cos²Θ - sin²Θ = sinΘ
1 - 2sin²Θ = sinΘ
0 = 2sin²Θ + sinΘ - 1
0 = (2sinΘ - 1)(sinΘ + 1)
sinΘ = 1/2, sinΘ= -1
Θ = π/6
Θ = 5π/6
Θ = 3π/2
for 0 ≤ Θ ≤ 2π