Solve. Between [0, 2pi] 2cos^2x+7cosx+3=0
1 answer:
9514 1404 393
Answer:
x = {2π/3, 4π/3}
Step-by-step explanation:
The equation can be factored as ...
(cos(x) +3)(2cos(x) +1) = 0
The solutions are the values of x where the factors are zero.
cos(x) +3 = 0
cos(x) = -3 . . . . . . no real solution
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2cos(x) +1 = 0
cos(x) = -1/2
x = arccos(-1/2) = 2π/3
The other solution in the desired interval is ...
x = 2π -(2π/3) = 4π/3
The solutions are x∈ {2π/3, 4π/3}.
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The solutions to the inequality are (-∞, -3] or a ≤ -3 if the inequality −14a≥ 42 graph is shown in the picture.
<h3>What is inequality?</h3>It is defined as the expression in mathematics in which both sides are not equal they have mathematical signs either less than or greater than known as inequality.
We have an inequality:
−14a ≥ 42
-a ≥ 3
a ≤ -3
a ∈ (-∞, -3]
Thus, the solutions to the inequality are (-∞, -3] or a ≤ -3 if the inequality −14a≥ 42 graph is shown in the picture.
Learn more about the inequality here:
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