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Chapter 7: Q. 26 (page 505)

Solve each equation on 0鈮�胃<2蟺.

-3cos蟺2-胃=tan胃

Short Answer

Expert verified

The solutions are胃=0,1.911,蟺,4.373.

Step by step solution

01

Step 1. Given Information

We are given the equation -3cos蟺2-胃=tan胃and we need to find the solutions of the equation on the interval0鈮�胃<2蟺.

02

Step 2. Simplifying the equation

We know, cos蟺2-胃=sin胃.

鈭�-3cos蟺2-胃=tan胃鈬�-3sin胃=tan胃鈬�-3sin胃-tan胃=0鈬�-3sin胃-sin胃cos胃=0鈬�-3sin胃cos胃-sin胃cos胃=0

鈭�-3sin胃cos胃-sin胃=0鈬�-sin胃(3cos胃+1)=0

03

Step 3. Finding the solutions

We get

sin胃=0,3cos胃+1=0

In the interval [0,2蟺)we have

胃=0,蟺from sin胃=0

Again,

role="math" localid="1646493616744" 3cos胃+1=0鈬�3cos胃=-1鈬�cos胃=-13鈬�胃=cos-1-13鈬�胃=1.911,4.373

Here, all the angles are in radian.

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Most popular questions from this chapter

Establish each identity.

2asin胃cos胃2+a2cos2胃-sin2胃2=a2

express each product as a sum containing only sines or only cosines

sin3蠎2cos蠎2

If z=tan伪2, show that localid="1646416257885" cos伪=1-z21+z2.

Multiply and simplify:(1+tan(胃))(1+tan(胃))-sec(胃)tan(胃)

Rewrite in terms of sine and cosine functions: cot(胃)sec(胃)
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