On which interval is y = sin(x) strictly increasing?

A function is strictly increasing in the interval where the derivate is positive.

Then we must look at the derivate of sin(x)

if f(x) = sin(x)

f'(x) = cos(x)

Then we must see which vales make cos(x) positive.

cos(0) = 1.

cos(90°) = 0

cos(-90°) = 0.

Then the cos(x) function is positive in the range (-90°, 90°) (notice that it is an open interval).

Then the function sin(x) will be strictly increasing in the interval (-90°, 90°)

And because these functions are periodic, where the period is 360°, we can actually write this interval as:

(-90° + n*360°, 90° + n*360°) where n is an integer number.

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

i think its a and d

step-by-step explanation: