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Chapter 5: Problem 29

Find the smallest positive number \(x\) such that $$ \sin \left(x^{2}+x+4\right)=0. $$

Short Answer

Expert verified
The smallest positive number \(x\) such that \(\sin \left(x^{2}+x+4\right)=0\) is \(x = (\sqrt{17} - 1)/2\).

Step by step solution

01

To find when the sine function is equal to 0, we need to recall the unit circle and the definition of sine. From the unit circle, we know that the sine of an angle is equal to the y-coordinate of the point where the terminal side of the angle intersects the unit circle. The sine function is equal to 0 at the points where the terminal side of the angle is on the x-axis. These angles are \(0\), \(\pi\), \(2\pi\) and so on. In general, any integer multiple of \(\pi\) will result in sine being equal to 0, i.e., $$\sin(n\pi)=0$$ where \(n\) is an integer. #Step 2: Set up the equation for the angle inside the sine function#

We want to determine a value of \(x\) such that the angle inside the sine function, \(x^2 + x + 4\), is an integer multiple of \(\pi\). Hence, we can set up an equation to represent this relationship: $$x^2 + x + 4 = n\pi$$ where \(n\) is an integer. #Step 3: Rearrange the equation to find x#
02

Rewrite the equation to solve for x: $$x^2 + x = n\pi - 4$$ As we need the smallest positive value for \(x\), we should try consecutive integer values of \(n\) and test if the equation yields a positive value for x. We can solve it using the quadratic formula: $$x = \frac{-1 \pm \sqrt{1^2-4(1)(n\pi - 4)}}{2(1)}$$ >x = \frac{-1 \pm \sqrt{1-4n\pi + 16}}{2} #Step 4: Check for the smallest positive value of x at consecutive integer values of n#

Now, we simply plug in consecutive integer values for \(n\) to find the smallest positive value of x. For \(n=0\), the equation becomes $$x = \frac{-1 \pm \sqrt{17}}{2}$$ The positive solution for \(x\) in this case is \(x=(\sqrt{17}-1)/2\), which is greater than 0. Therefore, the smallest positive number \(x\) such that $$ \sin \left(x^{2}+x+4\right) = 0 $$ is \(x = (\sqrt{17} - 1)/2\).

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