SOLUTION: Solve sin^2 x + sin x + 4 = 0 for x.

Question 1208766: Solve sin^2 x + sin x + 4 = 0 for x.
Found 2 solutions by Shin123, math_tutor2020:
Answer by Shin123(626) (Show Source): You can put this solution on YOUR website!
.

Let's consider the range of the sine function. must always be between -1 and 1.

This means that must always be between 0 and 1.

Combining these two means that must always be between -1 and 2.

Finally, adding 4 means must be between 3 and 6.

Therefore, there are no solutions. (assuming x is real)

Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Let w = sin(x)

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.

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Another way to look at it.

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.

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