Given the inequality:
sin(2x+4π)≥4sin(3π)To solve this inequality, we must first solve the corresponding equation:
sin(2x+4π)=4sin(3π)Solve:
Given the equation
sin(2x+4π)=4sin(3π)- this is the simplest trigonometric equation
This equation is transformed to
2x+4π=2πn+asin(0)2x+4π=2πn−asin(0)+πOr
2x+4π=2πn2x+4π=2πn+π, where n - is a integer
Move
4πto right part of the equation
with the opposite sign, in total:
2x=2πn−4π2x=2πn+43πDivide both parts of the equation by
2x1=πn−8πx2=πn+83πx1=πn−8πx2=πn+83πThis roots
x1=πn−8πx2=πn+83πis the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0≤x1For example, let's take the point
x0=x1−101=
(πn−8π)+−101=
πn−8π−101substitute to the expression
sin(2x+4π)≥4sin(3π)sin(2(πn−8π−101)+4π)≥4sin(3π)sin(-1/5 + 2*pi*n) >= 0
but
sin(-1/5 + 2*pi*n) < 0
Then
x≤πn−8πno execute
one of the solutions of our inequality is:
x≥πn−8π∧x≤πn+83π _____
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