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Solve each trigonmetric equation in the interval (0, 2pi) by first squarring both sides.

 

(SQ RT 3) sin x=cos x+1

 

PLS HELP W THE ANSWER

SOLUTION SET=????

 May 14, 2020
 #1
avatar+111456 
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√3sin x  =  cos x+ 1      square both sides

 

3sin^2 x   =  cos^2x  + 2cos x  + 1

 

3 ( 1 - cos^2x)  = cos^2x  + 2cosx  + 1

 

3 - 3cos^2 x = cos^2x + 2cos x + 1        rearrange as

 

4cos^2 x + 2cos x - 2  =  0         divide through by 2

 

2cos^2 x + cos x - 1  =  0     factor as

 

(2cos x - 1) ( cos x + 1)  =  0

 

So either

 

2cos x - 1  = 0                              

 

2cos x  = 1                                                            

 

cos x = 1/2                                                                

And this is true at  x = 60° and x =  360 - 60 = 300°

 

or

 

cos x + 1  =  0

 

cos x  = - 1

And this is true at x = 180°

 

Because we squred both sides  we need to check these solutions

 

√3 sin (60)  =  1 + cos (60) 

√3 (√3/2)  =  1 + 1/2

3/2  = 3/2      true

 

√3 sin (180)  = 1 + cos (180)

0 =  1  - 1

 0 = 0   true

 

√3 sin (300)  = 1 + cos (300)

√3 (-1/2)  = 1 + 1/2

-√3/2  =  3/2     not true

 

So...the  only solutions  are  x  = 60°   and  x   = 180°

 

cool cool cool

 May 14, 2020

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