The original equation given to you becomes w^2 + w + 4 = 0
You should find the discriminant is d = b^2-4ac = 1^2 - 4*1*4 = -15
Because d < 0, it means w^2 + w + 4 = 0 has two complex number solutions of the form a+bi, where i = sqrt(-1)
This must mean that the original equation also has complex number solutions of the form a+bi.
Therefore, sin^2(x)+sin(x)+4 = 0 does not have any real number solutions.
Here's a graph of y = ( sin(x) )^2 + sin(x) + 4.
The curve never touches the x axis.
Find the vertex of w^2 + w + 4. I'll let the student handle the scratch work, but you should find the lowest point for w^2 + w + 4 is when w = -1/2
Plug w = -1/2 into w^2 + w + 4 and you'll get 15/4 = 3.75
This shows that the smallest w^2 + w + 4 can get is 3.75, which unfortunately is above 0.
Meaning that w^2 + w + 4 will never be 0 no matter what you pick for w.
Consequently, no matter what real number you pick for x, sin^2(x)+sin(x)+4 will never be 0 either.