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Trigonometric Equation
- Thread starter TONY2014
- Start date
D
Deleted member 4993
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Could someone please help me? I'm trying to solve the following trig equation: 2cos(2x) -sin(x/2) = 0 over the interval 0 <x < pi. I keep getting stuck when I reach the point where I have 8[( 2cos2(x) - 1)2] = 1-cosx.
can you please show us how you got to:
8[( 2cos2(x) - 1)2] = 1-cosx
I don't get there!!
I apologize for my earlier post.
Here is my work:
2cos(2x) – sin(x/2) = 0
2cos(2x) = sin(x/2)
2cos(2x) = ± sqrt([1 − cos (x)] / 2)
(2cos(2x))2 = (1 – cos(x))/2
4cos2(2x) = (1-cos(x))/2
8cos2(2x) = 1 –cos(x)
8((2cos2 – 1)2) = 1 –cos(x)
8 (4cos4 (x) – 4cos2(x) + 1) = 1 –cos(x)
Here is the point at which I am confused.
Here is my work:
2cos(2x) – sin(x/2) = 0
2cos(2x) = sin(x/2)
2cos(2x) = ± sqrt([1 − cos (x)] / 2)
(2cos(2x))2 = (1 – cos(x))/2
4cos2(2x) = (1-cos(x))/2
8cos2(2x) = 1 –cos(x)
8((2cos2 – 1)2) = 1 –cos(x)
8 (4cos4 (x) – 4cos2(x) + 1) = 1 –cos(x)
Here is the point at which I am confused.
Last edited:
Here is my work:
2cos(2x) – sin(x/2) = 0
2cos(2x) = sin(x/2)
2cos(2x) = ± sqrt([1 − cos (x)] / 2)
(2cos(2x))2 = (1 – cos(x))/2
4cos2(2x) = (1-cos(x))/2
8cos2(2x) = 1 –cos(x)
8((2cos2 – 1)2) = 1 –cos(x)
8 (4cos4 (x) – 4cos2(x) + 1) = 1 –cos(x)
Here is the point at which I am confused.
Hi Tony:
This approach is fine, but you'll eventually need a numerical process to approximate the solutions. :idea: If you're using technology for that, then you could save some time by approximating the solutions directly, from a zoomed plot.
Your work looks good, so far. Minor point regarding the ± above: it's not needed, as that identity gives the opposite root over the given interval.
At this point, you could make the substitution cos(x) = u.
You'll get: 8(4u^4 - 4u^2 + 1) = 1 - u
Expand the left-hand side, and then subtract the right-hand terms from each side. Now you have a 4th-degree polynomial in u.
Use your favorite numerical process to approximate this polynomial's roots {R1, R2, R3, R4}. They all lie between -1 and 1.
For each root, you have the relationship ROOT = cos(x). Use the inverse cosine function, to recover x.
Now, since you squared both sides of an equation (during your solution process), you need to be aware of extraneous solutions. That is, you'll need to test each of your candidates, to see which lead to x values within the given interval AND make the original equation a true statement.
Cheers :cool: