Solve each equation for all values of x in the real number system.

Question

Solve each equation for all values of x in the real number system.  If an exact solution does not exist, round your answers to the nearest hundredth.

 

1.  sin2x=sinx

 

2.  cos^2x+cos2x=-1

 

Please show all steps necessary :)

Answer 1

Hi Kayla,

 

Let's look at the first equation.

 

sin2x = sinx

2sinxcosx = sinx (trig identity)

2sinxcosx - sinx = 0 (subtraction)

sinx(2cosx - 1) = 0  (factor)

 

sinx = 0 or 2cosx -1 = 0 (zero product property)

 

x=sin^-10  or cosx = 1/2

 

x = 0, π, 2π..., which we can write kπ where k is any integer.

 

For x = cos^-11/2 = π/3, -π/3 ... (to list all solutions: π/3 + 2kπ, k any integer or  -π/3+ 2kπ, k any integer)

 

There are several different ways to represent the set of all solutions, but this gives you the idea.

 

For 2)

 

cos²x + cos2x = -1

cos²x + (cos²x -1) = -1  (trig identity; since cos already appears in the equation choose to use this flavor of cos2x)

2cos²x = 0  (add 1)

 

cos²x = 0 (division)

cosx = 0 (square root)

 

x= cos^-10 = π/2 and -π/2, but there are infinitely many more solutions. One way to write that is:

 

x = π/2 + kπ, k any integer

Answer 2

1. we will have to use a few trig identities for this one

2. sin(2x) = sin x

3. using the double-angle identity: 2sinxcosx = sin x

4. cosx = sinx/2sinx = 1/2

5. x = cos^-1(1/2) = π/3 or -π/3

6. therefore, the solution for #1 is π/3 or -π/3

7. if #2 is cos²x + cos(2x) = -1, then we use the double-angle identity again:

8. cos²x + (1 - 2sin²x) = -1

9. now, we use the Pythagorean theorem: 1 - sin²x + 1 - 2sin²x = -1

10. combine like terms, and subtract both sides by -2: -3sin²x = -3

11. sin²x = 1

12. (sin x)² = 1

13. sin x = ±1

14. hence, x = sin^-1(1) and x = sin^-1(-1)

15. the solution for #2 is π/2 or -π/2