Search
Ask a question
0

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

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 :)

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Doug C. | Very patient, enthusiastic math tutor; know how to teach!Very patient, enthusiastic math tutor; k...
5.0 5.0 (575 lesson ratings) (575)
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
Anthony B. | Award-Winning Tutor with 5+ Years of Professional ExperienceAward-Winning Tutor with 5+ Years of Pro...
4.9 4.9 (125 lesson ratings) (125)
1
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