- #1
Petkovsky
- 62
- 0
cos(4x) + [tex]\sqrt{2}[/tex]cos(3x) + cos(2x) = 0
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Ok, so break down everything to a double angle and i get:
2cos^2(2x) - 1 + sqrt(2)*(sqrt((cos(2x) + cos^2(2x))/2) - sqrt((1-cos^(2x)*(1-cos(2x))/2) + cos(2x) = 0
...which is quite complicated to solve even if i substitute cos(2x) with 't'.
I also tried another aproach:
2cos^2(2x) - 1 + sqrt(2)*(cos(2x)cos(x) - sin(2x)(sin(x)) + 2cos(2x) - 1 = 0
here i aimed to complete a square but i can't see how.
Can you please give me some advice on how to continue or use another method
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Ok, so break down everything to a double angle and i get:
2cos^2(2x) - 1 + sqrt(2)*(sqrt((cos(2x) + cos^2(2x))/2) - sqrt((1-cos^(2x)*(1-cos(2x))/2) + cos(2x) = 0
...which is quite complicated to solve even if i substitute cos(2x) with 't'.
I also tried another aproach:
2cos^2(2x) - 1 + sqrt(2)*(cos(2x)cos(x) - sin(2x)(sin(x)) + 2cos(2x) - 1 = 0
here i aimed to complete a square but i can't see how.
Can you please give me some advice on how to continue or use another method