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Detecmine all soluhons for
0 <= t <= 4
2pi cos(pi t)+2pi cos(2pi t)=0

Detecmine all soluhons for $0 \leqslant t \leqslant 4$ $2 \pi \cos (\pi t)+2 \pi \cos (2 \pi t)=0$

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Solve two-step linear inequalities

Full solution

Q. Detecmine all soluhons for

0 \leqslant t \leqslant 4

2 \pi \cos (\pi t)+2 \pi \cos (2 \pi t)=0

Write Equation: First, let's write down the equation we need to solve: $2\pi \cos(\pi t) + 2\pi \cos(2\pi t) = 0$ .
Simplify Equation: We can simplify the equation by dividing everything by $2\pi$ : $\cos(\pi t) + \cos(2\pi t) = 0$ .
Find Solutions for $\cos(\pi t)$ : Now, let's find the values of $t$ that satisfy the equation for each cosine term separately within the given interval.
Find Solutions for $\cos(2\pi t)$ : For $\cos(\pi t) = 0$ , the solutions are $t = \frac{1}{2}$ and $t = \frac{3}{2}$ since $\cos(\frac{\pi}{2}) = 0$ and $\cos(\frac{3\pi}{2}) = 0$ .
Find Common Solutions: For $\cos(2\pi t) = 0$ , the solutions are $t = \frac{1}{4}$ , $t = \frac{3}{4}$ , $t = \frac{5}{4}$ , and $t = \frac{7}{4}$ since $\cos(\frac{\pi}{2}) = 0$ and $\cos(\frac{3\pi}{2}) = 0$ , and we need to divide by $2$ to account for the $2\pi t$ .
Correct Mistake: Now we need to find the common solutions for $\cos(\pi t) + \cos(2\pi t) = 0$ within the interval $0 \leq t \leq 4$ .
Correct Mistake: Now we need to find the common solutions for $\cos(\pi t) + \cos(2\pi t) = 0$ within the interval $0 \leq t \leq 4$ .Checking the solutions, we see that $t = \frac{1}{2}$ and $t = \frac{3}{2}$ are common solutions because at these points, $\cos(\pi t) = 0$ and $\cos(2\pi t)$ is also $0$ .
Correct Mistake: Now we need to find the common solutions for $\cos(\pi t) + \cos(2\pi t) = 0$ within the interval $0 \leq t \leq 4$ . Checking the solutions, we see that $t = \frac{1}{2}$ and $t = \frac{3}{2}$ are common solutions because at these points, $\cos(\pi t) = 0$ and $\cos(2\pi t)$ is also $0$ . However, we made a mistake in the previous step; we should have considered the sum of the cosines, not just their individual zeros. Let's correct this.

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