# Please solve for x in the interval 0 degrees less than x less than or equal to 2pi: sin^2x - 5sinx=0?

Relevance
• 9 years ago

We can factor out the common factor of sin(x) to get sin(x){sin(x) - 5} = 0

Thus we have that either sin(x) = 0

or

sin(x) - 5 = 0

solving sin(x) = 0: This happens when x is 0 radians and pi radians. However 0 radians isn't in our domain so we'll go with 2pi instead so its pi and 2pi

Solving sin(x) - 5 = 0 gives sin(x) = 5 This has no solution since sine is always between -1 and 1 so our final answer is x = pi, 2pi

• nitesh
Lv 4
4 years ago

So the question is: y= |cos(x)| the place 0 <= x <= 2pi y' = ? between 0 to pi/2, cos(x) is postive between pi/2 to 3pi/2, cos(x) is damaging between 3/pi/2 to 2pi, cos(x) is effective. to be able to rewrite as: y = cos(x) the place 3pi/2 <= x <= pi/2 y = -cos(x) the place pi/2 < x < 3pi/2 We did this to do away with definitely the cost, now we can compute y'. y' = -sin(x) the place 3pi/2 <= x <= pi/2 y' = sin(x) the place pi/2 < x < 3pi/2