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Chapter 7: Q. 75 (page 465)

In Problems 57鈥� 80, solve each equation on the interval 0鈮�胃鈮�2蟺

3(1-cos胃)=sin2胃

Short Answer

Expert verified

The solution set of the given equation is 0

Step by step solution

01

Step 1. Given Information 

In the given problems we have to solve each equation on the interval 0鈮�胃鈮�2蟺

3(1-cos胃)=sin2胃

02

Step 2. The equation in its present form contains a cosine and sine. 

However, a form of the Pythagorean Identity, sin2胃+cos2胃=1, can be used to transform the equation into an equivalent one containing only cosines.

3-3cos胃=1-cos2胃

Subtract 1 and add cos2胃on both side

3-3cos胃-1+cos2胃=1-cos2胃-1+cos2胃2-3cos胃+cos2胃=0cos2胃-3cos胃+2=0

03

Step 3. Now the equation is cos2θ-3cosθ+2=0

Factor the equation.

cos2胃-(2+1)cos胃+2=0cos2胃-2cos胃-cos胃+2=0cos胃(cos胃-2)-1(cos胃-2)=0(cos胃-1)(cos胃-2)=0

04

Step 4. Use the Zero-Product Property.

cos胃-1=0orcos胃-2=0cos胃-1+1=0+1orcos胃-2+2=0+2cos胃=1orcos胃=2

05

Step 5. Solving each equation in the interval [0,2π], we obtain  

胃=0orNosolution

The solution set is 0

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Most popular questions from this chapter

Establish each identity.

sin胃+cos胃cos胃-sin胃-cos胃sin胃=sec胃csc胃

Establish each identity.

(secv-tanv)2+1cscv(secv-tanv)=2tanv

If x=2tan胃, express sin2胃as a function ofx.

Suppose that f and g are two functions with the same domain. Iff(x)=g(x) for every x in the domain, the equation is called a(n) ____. Otherwise, it is called a(n) _____ equation.

tan2胃-sec2胃
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