The number of solutions of sinθ+2sin2θ+3sin3θ+4sin4θ=10, 0<θ<π is0124

The given equation is sinθ + 2sin2θ + 3sin3θ + 4sin4θ = 10.

First, we need to understand that the maximum value of sinθ is 1 and the minimum value is -1. Therefore, the maximum value of 2sin2θ is 2, of 3sin3θ is 3, and of 4sin4θ is 4.

Adding these maximum values together, we get 1 + 2 + 3 + 4 = 10.

However, it's important to note that these maximum values occur at different values of θ. For sinθ to be 1, θ must be π/2. For 2sin2θ to be 2, 2θ must be π/2, so θ must be π/4. Similarly, for 3sin3θ to be 3, 3θ must be π/2, so θ must be π/6. And for 4sin4θ to be 4, 4θ must be π/2, so θ must be π/8.

Since these values of θ are all different, it's impossible for the equation to equal 10 at any single value of θ in the given range 0 < θ < π.

Therefore, the number of solutions to the equation in the given range is 0.