Question #202352

Use the inequality 5cos(2x)≤0

5cos⁡⁡(2x)≤0 on [−π,π]

[−π,π] to answer the following questions.

Part A: What are the zeros to the nearest hundredth?

Part B: What is the solution set for the inequality?

Select all correct answers for Part A, and select one answer for Part B. Use π≅3.14

π≅3.14 if needed.


A: x= −1/2π

A: x= −3/4π

A: x= 0π

A: x= −1/4π

A: x= 1/4π

A: x= 3/4π

A: x=−1/2π


B: (−3/4π,−1/4π)∪(1/4π,3/4π)

B: [−1/2π,0π]∪[0π,1/2π]

B: [−1/2π,1/2π]

B: [−3/4π,−1/4π]∪[1/4π,3/4π]

B: (−∞,−3/4π]∪[−1/2π,1/2π]∪[3/4π,∞)


Expert's answer

5cos⁡(2x)≤0=>cos⁡(2x)≤05\cos(2x)\leq 0=>\cos(2x)\leq 0

If x∈[−π,π],x\in[-\pi, \pi], then 2x∈[−2π,2π]2x\in[-2\pi, 2\pi]


2x∈[−3π2,−π2]∪[π2,3π2]2x\in[-\dfrac{3\pi}{2}, -\dfrac{\pi}{2}]\cup[\dfrac{\pi}{2}, \dfrac{3\pi}{2}]

x∈[−3π4,−π4]∪[π4,3π4]x\in[-\dfrac{3\pi}{4}, -\dfrac{\pi}{4}]\cup[\dfrac{\pi}{4}, \dfrac{3\pi}{4}]

5cos⁡(2x)=05\cos(2x)= 0

=>x1=−3π4,x2=−π4,x3=π4,x4=3π4=>x_1=-\dfrac{3\pi}{4}, x_2=-\dfrac{\pi}{4}, x_3=\dfrac{\pi}{4}, x_4=\dfrac{3\pi}{4}


=>x1≈−2.36,x2=−0.79,x3=0.79,x4=2.36=>x_1\approx-2.36, x_2=-0.79, x_3=0.79, x_4=2.36


A: x= −3/4π

≈−2.36\approx-2.36


A: x= −1/4π

≈−0.79\approx-0.79


A: x= 1/4π

≈−0.79\approx-0.79


A: x= 3/4π

≈2.36\approx2.36


B: [−3/4π,−1/4π]∪[1/4π,3/4π]




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