BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter T, Problem 8DDT
To determine

To find: The values of x.

Expert Solution

Answer to Problem 8DDT

The values of x are 0and5π3.

Explanation of Solution

Given:

The function is sin2x=sinx on the interval [0,π].

Calculation:

Consider the given function sin2x=sinx.

sin2x=sinx2sinxcosx=sinx(sin2x=2sinxcosx)2sinxcosxsinx=0sinx(2cosx1)=0

That is, either

sinx=0x=nπ,nZ

Or

2cosx1=0cosx=12=cos(±π3)

Therefore, the value is x=2nπ±π3,nZ.

Here x on the interval [0,π].

Substitute n=0 in x=2nπ±π3,

x=2×0×π±π3=0±π3=π3,π3

Here x=π3 does not belongs to the interval [0,π].

Therefore, the values of x is π3.

Substitute n=1 in x=2nπ±π3,

x=2×1×π±π3=2π±π3=7π3,5π3

Here x=7π3 does not belongs to the interval [0,π].

Therefore, the values of x is 5π3.

Substitute n=2 in x=2nπ±π3,

x=2×2×π±π3=4π±π3=13π3,11π3

Here x=13π3,11π3 does not belongs to the interval [0,π].

Substitute n=1 in x=2nπ±π3,

x=2×(1)×π±π3=2π±π3=5π3,7π3

Here x=5π3,7π3 does not belongs to the interval [0,π].

Thus, the values of x are 0and5π3.

Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!