Answer:
its C
Step-by-step explanation:
sin(2x)-1 <=cos(2x)-2
whoever said it was wrong is looking at the wrong question.
The trigonometric inequalities over the interval [tex]0\leq \pi \leq 2[/tex] radians are [tex]\rm sin2x-1 = cos2x-2[/tex].
We have to determine, the following trigonometric inequalities over the interval [tex]0\leq \pi \leq 2[/tex] radians?
The trigonometric inequality over the interval [tex]0\leq \pi \leq 2[/tex] radians is used to solve the equation is as follows.
The equation is used to inequality over the given interval is,
[tex]\rm sin2x-1 = cos2x-2[/tex]
To solve the inequality substitute the formula of sin2x and cos2x.
[tex]\rm Sin2x = 2 \ sinx \ cosx\\\\Cos2x = 1-sin^2x[/tex]
Then,
The trigonometric inequality is,
[tex]\rm Sin2x-1 \leq cos2x-2\\\\\rm 2\ sinx \ cosx-1\leq 1-sin^2x - 2\\\\ 2\ sinx \ cosx-1 \leq -sin^2x - 1\\\\Adding \ 1 \ on \ both \ sides\\\\2\ sinx \ cosx-1 +1\leq -sin^2x - 1+1\\\\2\ sinx \ cosx-\leq -sin^2x \\\\2cosx \leq -sinx\\\\Divided\ by -cosx \ on \ both\ side\\\\2 \dfrac{cosx}{-cosx} \leq \dfrac{-sinx}{-cosx}\\\\tanx\leq 2[/tex]
Hence, The trigonometric inequalities over the interval [tex]0\leq \pi \leq 2[/tex] radians are [tex]\rm sin2x-1 = cos2x-2[/tex].
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For more details refer to the link given below.
https://brainly.com/question/18945214
