The solution of x for the trigonometric inequality [tex]2sin(x)+3 > sin^2(x)[/tex] is x will lie between 0≤x< [tex]\frac{3\pi}{2}[/tex]
What is trigonometric inequality?
An inequality of the standard form R(x) > 0 (or < 0) that consists of 1 or a few trigonometric functions of the variable arc x is a trigonometric inequality. Finding a solution to the inequality means determining the values of the variable arc x whose trigonometric functions make the inequality true.
According to the question
The trigonometric inequality :
[tex]2sin(x)+3 > sin^2(x)[/tex] and the interval 0<=x<=2pi radians
Let sin(x) = t
Now, inequality is :
[tex]2t+3 > t^{2}[/tex]
0 > t² - 2t - 3
0> t² - (3-1)t - 3
0> t² - 3t + t - 3
0> t( t - 3) + 1(t - 3)
0> ( t - 3) (t+1)
t < -1 or t < 3
sin(x) < -1 or sin(x) < 3
as sin(x) can not be 3 because sin oscillates between 1 and -1. Sine will never result in 3 so we disregard this as a solution.
sin(x) < -1
sin(x) < sin ([tex]\frac{3\pi}{2}[/tex] )
x < [tex]\frac{3\pi}{2}[/tex]
And the given interval for x
0<=x<=2pi radians
Therefore,
x will lie between
0≤x< [tex]\frac{3\pi}{2}[/tex]
Hence, The solution of x for the trigonometric inequality [tex]2sin(x)+3 > sin^2(x)[/tex] is x will lie between 0≤x< [tex]\frac{3\pi}{2}[/tex]
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