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How can we answer this question?

 

Guest Mar 4, 2018
 #1
avatar+7266 
+2

1.   sin2(x) - 2  =  -sin(x)

 

This can't be an identity because it is not always true. For instance...

 

sin2(0) - 2  =  -sin(0)

 

02 - 2  =  -0

 

-2  =  0        This is false, so this equation is not an identity.

 

However, this equation does have solutions. We can find its solutions.

 

sin2(x) - 2  =  -sin(x)

                                             Add  sin(x)  to both sides of the equation.

sin2(x) + sin(x) - 2  =  0

                                             Factor the left side like this:  u2 + u - 2  =  (u + 2)(u - 1)

(sin(x) + 2)(sin(x) - 1)  =  0

                                                             Set each factor equal to zero.

sin(x) + 2  =  0     or     sin(x) - 1  =  0

 

sin(x)  =  -2          or     sin(x)  =  1

 

sin  returns values between  -1  and  1  inclusively, so we can rule out the first option. That leaves...

 

sin(x)  =  1

 

x  =  arcsin(1)  =  90°

 

There will also be a solution every time  360°  is added to  90° , so all the real solutions are

 

x  =  90° + 360°n      where  n  is an integer

 

Does this answer your question? I might have misunderstood your question.

hectictar  Mar 4, 2018
edited by hectictar  Mar 4, 2018
 #2
avatar+7266 
+2

2.   (tan2x) / (1 + sec x)   =   sec x + 1      Let's test whether this is true when  x = 0 .

 

(tan2x) / (1 + 1/cosx)   =   1/cosx + 1

 

(tan20) / (1 + 1/cos0)   =   1/cos0 + 1

 

(02) / (1 + 1/1)   =   1/1 + 1

 

0  =  2       This is false, so this equation is not true for all values of  x  and so it is not an indentity.

 

If you want to find the solutions to the equation, we can try to find them.

 

(tan2x) / (1 + sec x)   =   sec x + 1       Multiply both sides by  (1 + sec x)

 

tan2x   =   (sec x + 1)(1 + sec x)

 

tan2x  =  sec x + sec2x + 1 + sec x

 

tan2x  =  sec2x + 2sec x + 1                Rewrite  tan  and  sec  in terms of  sin  and  cos.

 

sin2x / cos2x  =  1/cos2x + 2/cos x  +  1        Multiply through by  cos2x

 

sin2x  =  1 + 2cos x + cos2x                 Subtract  sin2x  from both sides.

 

0  =  1 + 2cos x + cos2x - sin2x           Substitute  1 - cos2x  in for  sin2x

 

0  =  1 + 2cos x + cos2x - (1 - cos2x)

 

0  =  1 + 2cos x + cos2x - 1 + cos2x

 

0  =  2cos x + 2cos2x         Divide through by  2 .

 

0  =  cos x + cos2x             Factor  cos x  out of both terms.

 

0  =  cos x(1 + cos x)         Set each factor equal to zero.

 

cos x  =  0      or      1 + cos x  =  0

 

However, neither of these can be true.

 

If cos x  =  0  then  sec x  is undefined and the original equation is undefined.

 

If  cos x  =  -1  then  1 + sec x  =  1 - 1  =  0  and the original equation is undefined.

 

So this equation has no solutions.

hectictar  Mar 4, 2018

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