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Problem 921

# Find the solution set on $$[0,2 \pi]$$ of the equation $\left(\sqrt{1}+\sin ^{2} x\right)=(\sqrt{2}) \sin x$.

Expert verified
The solution set for the equation $$\left(\sqrt{1}+\sin ^{2} x\right)=(\sqrt{2}) \sin x$$ on the interval $$[0,2 \pi]$$ is empty, as there are no real solutions for $$y = \sin x$$.
See the step by step solution

## Step 1: Simplify the equation

First, notice that $$\sqrt{1}=1$$, so the equation can be written as: $\sin^2x + 1 = \sqrt{2}\sin x$

## Step 2: Rearrange the equation to a quadratic form

In order to solve this equation for x, let's try to write it in a quadratic form. Subtract $$\sqrt{2}\sin x$$ from both sides, obtaining: $\sin^2x - \sqrt{2}\sin x + 1 = 0$ Now, let $$y = \sin x$$, so the equation becomes a quadratic equation in terms of y: $y^2 - \sqrt{2}y + 1 = 0$

## Step 3: Solve the quadratic equation

To find the roots of the quadratic equation $$y^2 - \sqrt{2}y + 1 = 0$$, we can use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ In this case, $$a=1, b=-\sqrt{2},$$ and $$c=1$$. Plugging these values into the formula, we get: $y = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2-4(1)(1)}}{2}$ Evaluating the expression under the square root, we find that it is negative: $(-\sqrt{2})^2-4(1)(1) = 2 - 4 = -2$ Since we have a negative number under the square root, there are no real roots for the quadratic equation, so there are no solutions for $$y = \sin x$$.

## Step 4: Conclusion

As there are no real solutions for $$y = \sin x$$, this means there are no values of x in the interval $$[0, 2\pi]$$ that make the equation $$\left(\sqrt{1}+\sin ^{2} x\right)=(\sqrt{2}) \sin x$$ true. Therefore, the solution set for the equation is empty.

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