Asked by sqleung
Solve for x (exact solutions):
sin x - sin 3x + sin 5x = 0
-¦Ð ¡Ü x ¡Ü 0
------
Now, using my graphics calculator, I discovered that the following equation has 5 solutions. I have managed to come up with 3 of them but I'm having trouble finding the other 2. My 3 answers are:
0, -¦Ð/6, -5¦Ð/6
I need the answers to be EXACT like above so any help would be appreciated. Thankyou.
sin x - sin 3x + sin 5x = 0
-¦Ð ¡Ü x ¡Ü 0
------
Now, using my graphics calculator, I discovered that the following equation has 5 solutions. I have managed to come up with 3 of them but I'm having trouble finding the other 2. My 3 answers are:
0, -¦Ð/6, -5¦Ð/6
I need the answers to be EXACT like above so any help would be appreciated. Thankyou.
Answers
Answered by
drwls
I answered this question two days ago. Once of the answers is of course x = 0. For the others, I substitued indentities for sin 3x and sin 5x in terms of sin x, and ended up with a factorable equation for sin x. This leads to exact equations for arcsin in terms of fractions.
I am not able to read your terms -¦Ð/6 and -5¦Ð/6
If I can find my original answer, I will post a link to it below
I am not able to read your terms -¦Ð/6 and -5¦Ð/6
If I can find my original answer, I will post a link to it below
Answered by
drwls
OK, I found my original answer at
http://www.jiskha.com/display.cgi?id=1204243109
and it has two mistakes, a missing exponent and an incorrect sign. I should have ended up with the equation
3sinx - 16 sin^3x + 16sin^5x = 0
which factors to
sinx (3 - 16sin^2x + 16sin^4x) = 0
sinx (4sin^2x-3)(4sin^2x -1) = 0
Thge roots, besides x=0, are:
sin^2x = 1/4
sin x = +/-sqrt(1/2)
and
sin^2 = (3/4)
sin x = +/-(sqrt3)/2
http://www.jiskha.com/display.cgi?id=1204243109
and it has two mistakes, a missing exponent and an incorrect sign. I should have ended up with the equation
3sinx - 16 sin^3x + 16sin^5x = 0
which factors to
sinx (3 - 16sin^2x + 16sin^4x) = 0
sinx (4sin^2x-3)(4sin^2x -1) = 0
Thge roots, besides x=0, are:
sin^2x = 1/4
sin x = +/-sqrt(1/2)
and
sin^2 = (3/4)
sin x = +/-(sqrt3)/2
Answered by
drwls
Any positive or negative integral multiple of 30 degrees (x=pi/6), including 0 degrees, satisfies the equation.
Answered by
drwls
All +/- integral multiples of pi/6 EXCEPT multiples of pi/2 are solutions
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