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Chapter 29: Chapter 29

Problem 921

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

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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 \(y = \sin x\).
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TABLE OF CONTENTS :
TABLE OF CONTENTS

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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