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# Show that the equation has exactly one real root.$2x + \cos x = 0$

## Let $f(x)=2 x+\cos x .$ Then $f(-\pi)=-2 \pi-1<0$ and $f(0)=1>0 .$ since $f$ is the sum of the polynomial $2 x$ and thetrignometric function $\cos x, f$ is continuous and differentiable for all $x$. By the Intermediate Value Theorem, there is a number$c$ in $(-\pi, 0)$ such that $f(c)=0 .$ Thus, the given equation has at least one real root. If the equation has distinct real roots $a$ and$b$ with $a<b,$ then $f(a)=f(b)=0 .$ since $f$ is continuous on $[a, b]$ and differentiable on $(a, b),$ Rolle's Theorem implies thatthere is a number $r$ in $(a, b)$ such that $f^{\prime}(r)=0 .$ But $f^{\prime}(r)=2-\sin r>0$ since $\sin r \leq 1 .$ This contradiction shows that thegiven equation can't have two distinct real roots, so it has exactly one root

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Okay. Uh We want to solve this equation. Two X plus co sign X equals zero. So a solution to this equation is a root of the equation. When you have an equation equals zero. The solution is called a route of the equation. Now. Uh We're actually going to look at this graphically but first um I want to show you that uh if we subtracted two X from both sides of the equation we would have co signed X equals two X. No co sign the co sign function alternates between one and negative one. It takes on all the values between one and negative one. But if you remember the graph of a co sign function, it oscillates up and down. It never goes above one and it never goes below negative one. So two X would actually be bounded um between one and negative one. The reason for this is if two X is equal to cosign effects. Actually uh if I subtracted, if I saw it for co sign next, I would have to subtract two X from both sides. So Co sign X equals negative two X. No big difference. Um If I subtracted two X from both sides, I get cosign X equals negative two X. Because co sign X is bounded between one and negative one, -2 X would be bounded between one and negative one. And if we solve for X in this little inequality dividing everything by negative two we would get X itself has to be between one half and negative one half. We're still gonna solve ah this equation graphically but I just wanted to show you um that we really only need to look at the values of X between negative one half and one half. Uh For example, suppose we let X be out of this range. For example, X was three Effects was three negative two times X would be negative two times three, which is negative six. Co sign of X is never going to equal negative six. Co sign does not go above one, does not go below a negative one. It's not going to equal negative six. And that's why X would be restricted to this interval. So now let's use the dez most graphing calculator. And let's look at this function two X plus Co sign effects. Uh in the interval where X is between negative one half and one half. And see if we can find a solution to this equation. Now, a solution to this equation, a root of this equation will be a value of X. Uh Where this function equal zero. If we can find a value of X that makes this function zero dysfunction will be equal to zero when the graph crosses the X axis. Okay, because the function value is the y coordinate. And if you want the function value or the y coordinate to equal zero. We're looking for the X intercept. Let's look at the graph of two X plus cosign X. Two X plus co sign of X. Here is the graph of two X plus co sign X. You can see that it has an X intercept right here at this point. So when x is negative 0.45, the function two X plus co sign of X equals zero. So X equals negative 0.5 is a solution to this equation equaling zero when X is negative 0.45. This equation equals zero. Now you can see remember how I said, we only want to look at excess between one half and negative one half. We don't really have to worry about what the graph does as it moves on to the right or as it moves on to the left between X equals negative one half and positive one half. You can see that this graph across the D X. Axis only once. So there is only one route Or one solution to the equation. So two X plus co sign of X equals zero. When X equals -0.45. When x equals negative 0.45 we have a solution to this equation. Two X plus co sign of X equals zero. Uh this was the only solution. And so this equation has only one solution or one real numbered route and that is when x equals negative 0.45

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