5. Let h(x)=3cosx and let k(x)=1+5cosx . What are all values of x in the xy-plane, 0leqslant xleqslant 2pi for which h(x)leqslant k(x)

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5. Let h(x)=3cosx and let k(x)=1+5cosx . What are all values of x in the xy-plane, 0leqslant xleqslant 2pi for which h(x)leqslant k(x)

Answer 1

$x \in [0, \frac{\pi}{3}] \cup [\frac{5\pi}{3}, 2\pi] $

Answer 2

To figure out where $h(x) \leq k(x)$, we need to solve the inequality. Let's start by setting $h(x) = k(x)$, or $3cosx = 1 + 5cosx$. Solving this for $cosx$ gives us $cosx = \frac{1}{2}$. The inequality $h(x) \leq k(x)$ means that the cosine curve for $3cosx$ is below or equal to $1+5cosx$. Next, by examining the nature of the cosine function, which fluctuates between -1 and 1 regardless of the scalar multiply (the 3 and the 5 here), both curves oscillate equally, but $k(x)$ will always be 1 unit above. The points of intersection occur when $cosx = \frac{1}{2}$. The solutions to this lie in the given interval, $0 \le x \le 2\pi $ are $x= \frac{\pi}{3}$ and $x= \frac{5\pi}{3}$ from the nature of cosine function. However the feeling inequality points to $h(x) \le k(x)$ so consider values less than these points or in between of such occurrences. Thus, in the interval $0 \le x \le 2 \pi$, $h(x) = 3cosx \le 1+5cosx = k(x)$ for interval set given by: Time for cosx = 1/2: $x_1$ = 1/cos^-1[ 1/2], and $x_2$ = (2pi — $x_1$). Always less than equal TIme for g(x) > h(x) only twice a integral hemisphere of cosine so it is present only between these times above

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5. Let h(x)=3cosx and let k(x)=1+5cosx . What are all values of x in the xy-plane, 0leqslant xleqslant 2pi for which h(x)leqslant k(x)

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