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Sketch the graphs of each of the following for $x$ in [0,2$\pi$]. list the $x$-axis intercepts of each graph for this interval.

$$y=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1$$ I tried to solve the equation $y = 0$ by performing the following steps: $$\begin{align*} -1&=\sqrt{2} \cos\left(x-\frac{\pi}{4}\right) \\\\ -\frac{1}{\sqrt{2}}&=\cos\left(x-\frac{\pi}{4}\right) \\\\ -\frac{1}{\sqrt{2}}+\frac{\pi}{4}&=\cos(x) \end{align*}$$

I'm unable to proceed further. Can you please explain this to me in a step-by-step fashion? And I think that after you've got the answer you have to look the unit circle to get the exact answer. Please also show me how to look for the answer from unit cirle.

Thanks so much!

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Note that $\cos(x - \pi/4) \neq \cos(x) - \pi/4$. Hint: do you know the solutions to $\cos x = 1/\sqrt{2}$? What is $\cos (x+\pi)$? – Martin Wanvik May 8 '12 at 11:57
yea! i think $cosx$=$\frac{\sqrt{2}}{\sqrt{2}}$ and its show in the unit cirle $\frac{7\pi}{4}$ – Sb Sangpi May 8 '12 at 12:06
Not sure I understand - yes, you are correct in that one solution to $\cos x = -1/\sqrt{2}$ is $7\pi/4$, but there is also another one (on, say, the interval $[0,2\pi)$) If you're having difficulties, I suggest that you draw a unit circle, determine where the angle $7\pi/4$ is located, and project that point on the $x$-axis to get the cosine of the angle - then it should be easy to find another angle whose cosine is the same. Then it is a simple matter of taking each solution $y$ of $\cos y = -1/\sqrt{2}$ and solving $x - \pi/4 = y$ for $x$. – Martin Wanvik May 8 '12 at 12:17
Note that $y=\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+1=\cos(x)+\sin(x)+1$. – Jon May 8 '12 at 12:53
up vote 2 down vote accepted

As mentioned in the comments, you can't add $\frac{\pi}{4}$ to both sides, you can think of it as being "trapped" inside the cosine. Your work is correct up to your second step: \begin{align*} -1&=\sqrt{2} \cos\left(x-\frac{\pi}{4}\right) \\\\ -\frac{1}{\sqrt{2}}&=\cos\left(x-\frac{\pi}{4}\right) \end{align*} Now, this is when you look at the unit circle. You will notice that $\cos(x) = -\frac{1}{\sqrt{2}}$ when $x = \frac{3\pi}{4}$ and $x = \frac{5\pi}{4}$. This means that we have to solve the following equations: $$ x-\frac{\pi}{4} = \frac{3\pi}{4} \quad \quad \mbox{and} \quad \quad x-\frac{\pi}{4} = \frac{5\pi}{4}. $$ Doing this gives you $x = \pi$ and $x = \frac{3\pi}{2}$.

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Let $S=\left\{z \in \mathbb{R} \mid \cos z = -\frac{\sqrt{2}}{2}\right\}$. Prove that $x$ solves $\sqrt{2}\cos \left(x-\frac{\pi}{4}\right)+1=0$ if and only if $x-\frac{\pi}{4} \in S$. Finally, pick only the $x$'s belonging to $[0,2\pi]$.

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