# Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing. - Mathematics

Find the intervals in which f(x) = sin 3x – cos 3x, 0 < x < π, is strictly increasing or strictly decreasing.

#### Solution

Consider the function f'(x) = sin3x - cos3x.

f'(x)= 3cos3x+3sin3x

=3(sin3x + cos3x)

=3sqrt2{sin3xcos(pi/4)+cos3xsin(pi/4)}

=3sqrt(2){sin(3x+pi/4)}

For the increasing interval f'(x)>0

3sqrt2{sin(3x+pi/4)}>0

sin(3x+pi/4)>0

0<3x+π/4<π

=>0<3x<(3pi)/4

⇒ 0 < x < π/4

Also

sin(3x+pi/4)>0

when, 2pi<3x+pi/4<3pi

=>(7pi)/4<3x<(11pi)/4

Therefore, intervals in which function is strictly increasing in  0 < x < π/4 and 7π/12< x <11π/12.

Similarly, for the decreasing interval f'(x)< 0.

3sqrt2{sin(3x+pi/4)}<0

sin(3x+pi/4)<0

=>pi<3x+pi/4<2pi

=>(3pi)/4<3x<(7pi)/4

⇒ π/4 < x <7π/12

Also

sin(3x+pi/4)<0

When

3pi<3x+pi/4<4pi

=>(11pi)/4<3x<(15pi)/4

=>(11pi)/12

The function is strictly decreasing in pi/4and  `(11pi)/12

π4<x<7π12 and 11π12<x<π" data-mce-style="position: relative;" data-mce-tabindex="0">π4<x<7π12 and 11π12<x<π

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