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Chapter 7: Q. 28 (page 505)

Solve each equation on 0鈮�胃<2蟺.

sin(胃+1)=cos胃

Short Answer

Expert verified

The solutions are胃=0.285,3.427.

Step by step solution

01

Step 1. Given Information 

We are given the equation sin(胃+1)=cos胃 and we need to find the solutions of the equation on the interval 0鈮�胃<2蟺.

02

Step 2. Simplifying the equation 

We know, sin伪+尾=sin伪cos尾+cos伪sin尾.

鈭�sin胃+1=cos胃鈬�sin胃cos1+cos胃sin1=cos胃

Dividing both sides by cos胃we get,

tan胃cos1+sin1=1鈬�tan胃cos1=1-sin1鈬�tan胃=1-sin1cos1鈬�tan胃=sec1-tan1鈬�胃=tan-1sec1-tan1
03

Step 3. Finding the solutions

鈭�胃=tan-1sec1-tan1鈬�胃=tan-10.293鈬�胃=0.285

The tangent function has a period of 蟺, the solution is:

胃=0.285,0.285+2蟺鈬�胃=0.285,3.427

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