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# Pre-Calc Question

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4cos(x)-4sin^2(x) + 5=0

x = ?

Jan 12, 2018

#1
+8071
+1

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.

Jan 12, 2018

#1
+8071
+1

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.

hectictar Jan 12, 2018