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Trigonometric Laws and Identities

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What are the solutions to the trigonometric equation on the interval [0,2π)?

2cos^2x=cosx

Select all correct solutions.

0

π6

π3

π2

2π3

5π6

π

7π6

4π3

3π2

5π3

11π6

Jun 4, 2021

#2
+1

Need to be  careful  about  dividing out  possible solutions.........

2cos^2x   =  cos x

2cos^2 x  - cos x   =  0

cos  x ( 2cos x  - 1)  =  0

Set each factor to 0 ,  and solve for  x

cos x =  0        x =   pi/2      and  x = 3pi/2                           2cos x - 1   =  0

2cosx  =  1

cosx =  1/2      x = pi/3  and x = 5pi/3   Jun 4, 2021

#1
+1

2cos^2x=cosx       Divide both sides of th equation by cos

2 cos x = 1

cos x = 1/2

x = pi/3    5 pi / 3

Jun 4, 2021
#2
+1

Need to be  careful  about  dividing out  possible solutions.........

2cos^2x   =  cos x

2cos^2 x  - cos x   =  0

cos  x ( 2cos x  - 1)  =  0

Set each factor to 0 ,  and solve for  x

cos x =  0        x =   pi/2      and  x = 3pi/2                           2cos x - 1   =  0

2cosx  =  1

cosx =  1/2      x = pi/3  and x = 5pi/3   CPhill Jun 4, 2021
#3
+1

Thanx, Chris !    ~ EP

Jun 4, 2021