What is the solution to the trigonometric inequality sin(x)>1+cos(X) over the interval 0<=x<=2pi radians?

Answer:

Last choice

Step-by-step explanation:

First, convert (sinx)^2 to 1-(cosx)^2 via trigonometric identity derived from (sinx)^2 + (cosx)^2 = 1

[tex]\displaystyle \large{\sin^2x + \cos^2x=1 \longrightarrow \sin^2x=1-\cos^2x}[/tex]

Thus, the rewritten inequality is [tex]\displaystyle \large{1-\cos^2x>1+\cos x}[/tex]

Notice how the inequality has a quadratic pattern since we have (cosx)^2 and cosx in the inequality, simply arrange degrees in order.

[tex]\displaystyle \large{1-\cos^2x>1+\cos x}\\\displaystyle \large{0>1+\cos x-1+\cos^2x}\\\displaystyle \large{0>\cos x+\cos^2x}\\\displaystyle \large{\cos^2x+\cos x < 0}[/tex]

Then we factor cosx since the expression has cosx as a common factor.

[tex]\displaystyle \large{\cos x(\cos x+1)<0}[/tex]

Then we have to consider each inequalities which are cosx<0 and cosx+1<0 then merge both intervals once obtaining the solutions.

(1) [tex]\displaystyle \large{\cos x < 0}[/tex]

This inequality means that we have to find the x-value(s) that satisfy(ies) it.

Since cosx becomes negative after π/2 (Refer to unit circle, cosx becomes negative in second quadrant and this quadrant), that means it becomes negative until it reaches or < 3π/2.

Thus, [tex]\displaystyle \large{\frac{\pi}{2}<x<\frac{3\pi}{2}}[/tex] is the solution to cosx > 0.

The another way to solve inequality is to find the intersections by solving the equation then joint the interval together.

[tex]\displaystyle \large{\cos x<0 \longrightarrow \cos x=0}[/tex]

From the unit circle, cosx becomes 0 at x = π/2 and x = 3π/2 between interval [0,2π]

Therefore:

[tex]\displaystyle \large{\cos x = 0 \longrightarrow x = \frac{\pi}{2}, \frac{3\pi}{2}}\\\displaystyle \large{\boxed{\frac{\pi}{2}<x<\frac{3\pi}{2}}}[/tex]

(2) [tex]\displaystyle \large{\cos x+1<0 \longrightarrow \cos x <-1[/tex]

For this inequality, it appears that there have no solutions because cosine function is defined between [-1,1] in range hence there are no x-values in real plane that make cosx even less than -1.

Because only π satisfies cosx = -1 but not cosx < -1, we restrict π and every values.

Hence no solutions which cause a break of interval between π and π/2 < x < 3π/2

If you substitute x = π in, the inequality is false and therefore the answer is:

[tex]\displaystyle \large{\frac{\pi}{2}<x<\pi, \pi<x<\frac{3\pi}{2}}[/tex]

Answer:

D.

Step-by-step explanation:

The first two don't work, so it's between the last two.

This equation doesn't work for 180, so the answer is the last choice.

:)

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