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Chapter 7: Q. 75 (page 504)

Solve the equation on the interval 0鈮�胃鈮�2蟺.

sin胃-cos胃=1

Short Answer

Expert verified

On solving the equation, we get,

胃鈭�蟺2,蟺

Step by step solution

01

Step 1. Given information.

Consider the given question,

sin胃-cos胃=10鈮�胃鈮�2蟺

Squaring both sides of the equation,

sin胃-cos胃2=12sin2胃+cos2胃-2sin胃路cos胃=11-2sin胃路cos胃=12sin胃路cos胃=0

Then,

2sin胃=0sin胃=0

Also,

2cos胃=0cos胃=0

02

Step 2. Solve the equation for the given interval.

Consider the given question,

sin胃=0cos胃=00鈮�胃鈮�2蟺

Solving sin胃=0for the given interval,

sin胃=0胃=sin-10胃=0,蟺

Solving cos胃=0 for the given interval,

cos胃=0胃=cos-1=0胃=蟺2,3蟺2

03

Step 3. Check the solutions to see if any are extraneous.

As we squared both sides, we must check the solutions to see if any are extraneous,

If 胃=0, then,

=sin0-cos0=0-1=-1

If 胃=蟺, then,

=sin蟺-cos蟺=0--1=1

If 胃=蟺2, then,

=sin蟺2-cos蟺2=1-0=0

If 胃=3蟺2, then,

=sin3蟺2-cos3蟺2=-1-0=-1

Hence, we see 胃=0,3蟺2are extraneous.

Therefore, the solution set is胃鈭�蟺2,蟺

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