Pre-Calc Question

4cos x  -  4sin²x  +  5   =   0

                                                             From the Pythagorean identity,   sin²x   =   1 - cos²x

4cos x  -  4( 1 - cos²x )  +  5   =   0

                                                             Distribute the  -4  to both terms in parenthesees.

4cos x  -  4  +  4cos²x  +  5   =   0

                                                             Combine the  -4  and  +5   to get  +1 , and rearrange.

4cos² x  +  4cos x  +  1   =   0

 

4( cos x )²  +  4( cos x )  +  1   =   0     This is a quadratic equation which factors like this...

 

4( cos x )²  +  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   =   0

                                          Take the square root of both sides.

2( cos x ) + 1   =   0

                                          Subtract  1  from both sides.

2( cos x )   =   -1

                                          Divide both sides by  2 .

cos x   =   -1/2

 

x   =   120°   +   360n°

 

x   =   240°   +   360n°       where  n  is an integer.