Find all values of A in the interval 180deg less than or equal to A less than or equal to 360deg that satisfy sin(A) - sin(2A) =0

Question

It has to be solved algebraically. It also has to find all the values. 

Answer 1

Hy Ashley,

 

 

you have got to remember that Sin 2 A can be written as 2 Sin A Cos A

 

 

Now let's get to solving the equation

 

sin (A) - 2 sin (A) cos (A) = 0

 

 

sin (A) ( 1 - 2 Cos (A) ) = 0

 

Cos ( A) = 1/ 2 & Sin ( A) = 0

 

I'm going to give the values of A that satisfy Cos A = 1/2

 

arc cos (0.5) = 60 degrees

 

Cosine is positive in the first and last quadrants the only one we need to worry about is the last one thus

 

A = 360 - 60 

 

= 300°

 

We need not worry about sine as it is positive only in the first and second quadrants meaning it's not within our range for A which is 180 ≤ A ≤ 360.

 

Hope my feedback was helpful if it was give me a thumbs up .

 

Regards,

 

Mayuran

Answer 2

First, notice that sin(2A) is a double angle identity.

 

sin(2A) = 2sin(A)cos(A)

 

 

Therefore, you have

 

sin(A) - 2sin(A)cos(A) = 0

 

 

Factor the left side of the equation using by extracting the GCF.

 

sin(A)(1 - 2cos(A)) = 0

 

 

Set the factors equal to zero.

 

sin(A) = 0                      and                 1 - 2cos(A) = 0

 

     A = 0                                                 -2cos(A) = -1

     A = 180                                       

     A = 360                                                 cos(A) = 1/2

                

                                                                        A = 60

                                                                        A = 360 - 60 = 300

 

 

Out of all the solutions, select the ones that are in the given domain.  Then, verify them by plugging then into the original equation.