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For 0 < x < pi

4*sin2(x) = 4*cos(x) + 5

x = ?

 Jun 22, 2018
 #1
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4 sin2x   =   4 cos x  +  5

                                                   By the Pythagorean identity,   sin2x  =  1 - cos2x    so....

4( 1 - cos2x )   =   4 cos x  +  5

                                                   Distribute the  4  to the terms in parenthesees.

4  -  4 cos2x   =   4 cos x  +  5

                                                   Add   4 cos2 x   to both sides of the equation.

4   =   4 cos2x  +  4 cos x  +  5

                                                   Subtract  4  from both sides of the equation.

0   =   4 cos2x  +  4 cos x  +  1

                                                   Split   4 cos x   into two terms that we can use to factor by grouping.

 

0   =   4 cos2x  +  2 cos x  +  2 cos x  +  1

 

0   =   2 cos x ( 2 cos x  +  1 )  +  1( 2 cos x  +  1 )

 

0   =   ( 2 cos x  +  1 )( 2 cos x  +  1 )

 

0   =   ( 2 cos x  +  1 )2

                                      Take the square root of both sides.

0   =   2 cos x  +  1

                                      Subtract  1  from both sides.

-1   =   2 cos x

                                      Divide both sides by  2 .

-1/2   =   cos x

 

The angle that has a cosine of   -1/2   in the interval  [0, pi)   is   2 pi / 3

 

x  =  2 pi / 3

 Jun 22, 2018

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