Solve for x (exact solutions):
sin x - sin 3x + sin 5x = 0
-¦Ð ¡Ü x ¡Ü 0
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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.
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
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
Any positive or negative integral multiple of 30 degrees (x=pi/6), including 0 degrees, satisfies the equation.
All +/- integral multiples of pi/6 EXCEPT multiples of pi/2 are solutions
To solve the equation sin x - sin 3x + sin 5x = 0 for exact solutions, we can use trigonometric identities and algebraic manipulations.
1. Start by factoring out sin x from all three terms:
sin x - sin 3x + sin 5x = sin x(1 - 3cos 2x + 5cos^2 x - 1 + 3cos^2 x)
= sin x(8cos^2 x - 3cos 2x)
2. Now, we have the equation sin x(8cos^2 x - 3cos 2x) = 0. This equation will be satisfied if either sin x = 0 or 8cos^2 x - 3cos 2x = 0.
3. Solve sin x = 0:
In the given interval -π ≤ x ≤ 0, we know that x = 0 is a solution. To find the other solutions, we can use the fact that sin x = 0 when x is any integer multiple of π. So, we can write the general solution as x = nπ, where n is an integer.
4. Solve 8cos^2 x - 3cos 2x = 0:
To simplify this equation, we need to express cos 2x in terms of cos x. We can use the double angle formula: cos 2x = 2cos^2 x - 1.
Substitute cos 2x into the equation:
8cos^2 x - 3(2cos^2 x - 1) = 0
8cos^2 x - 6cos^2 x + 3 = 0
2cos^2 x + 3 = 0
We can solve this quadratic equation by factoring or using the quadratic formula. Since cos x is a trigonometric function, we know that the value is bounded between -1 and 1. Therefore, there are no real solutions for cos x in this case.
In conclusion, the exact solutions for the equation sin x - sin 3x + sin 5x = 0 in the interval -π ≤ x ≤ 0 are x = 0 and x = nπ, where n is an integer.