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sin 2x + 2(2 cos x - 3 sin x) =12

 Jul 12, 2016

Best Answer 

 #3
avatar+27476 
+15

A quick graph shows pro35hp is clearly correct:

 

 Jul 12, 2016
 #1
avatar+135 
+5

Subtract 12 from each side

will get \(\sin \left(2x\right)+2\left(2\cos \left(x\right)-3\sin \left(x\right)\right)-12=0\)

Use the identity \(\sin \left(2x\right)=2\cos \left(x\right)\sin \left(x\right)\)

\(\left(2\cos \left(x\right)-3\sin \left(x\right)\right)2+2\cos \left(x\right)\sin \left(x\right)-12=0\)

\(2\left(2\cos \left(x\right)+\cos \left(x\right)\sin \left(x\right)-3\sin \left(x\right)-6\right)\)

\(=2\left(\sin \left(x\right)+2\right)\left(\cos \left(x\right)-3\right)\)

solving each part equivalent to 0

\(\sin \left(x\right)+2=0\:\:\:\mathrm{or}\:\:\:\cos \left(x\right)-3=0\)

sinx cannot be -2

and cos x cannot be 3 

hence , no answer in R

 Jul 12, 2016
 #2
avatar
0

Solve for x:
2 (2 cos(x)-3 sin(x))+sin(2 x) = 12

 

2 (2 cos(x)-3 sin(x))+sin(2 x)  =  4 cos(x)-6 sin(x)+sin(2 x):
4 cos(x)-6 sin(x)+sin(2 x) = 12

 

Subtract 12 from both sides:
-12+4 cos(x)-6 sin(x)+sin(2 x) = 0

 

Simplify trigonometric functions:
2 (cos(x)-3) (2+sin(x)) = 0

 

Divide both sides by 2:
(cos(x)-3) (2+sin(x)) = 0

 

Split into two equations:
cos(x)-3 = 0 or 2+sin(x) = 0

 

Add 3 to both sides:
cos(x) = 3 or 2+sin(x) = 0

 

Take the inverse cosine of both sides:
x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z or x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  2+sin(x) = 0

 

Subtract 2 from both sides:
x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z
   or  x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  sin(x) = -2

 

Take the inverse sine of both sides:
Answer: |  x = cos^(-1)(3)+2 pi n_1  for  n_1  element Z
   or  x = 2 pi n_2-cos^(-1)(3)  for  n_2  element Z
   or  x = pi+sin^(-1)(2)+2 pi n_3  for  n_3  element Z                                                                           or x = 2 pi n_4- sin^(-1)(2)  for  n_4  element Z

 Jul 12, 2016
 #3
avatar+27476 
+15
Best Answer

A quick graph shows pro35hp is clearly correct:

 

Alan Jul 12, 2016

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