If I have a such equation to solve: sin⁡3x=12

guringpw

guringpw

Answered question

2021-12-26

If I have a such equation to solve:
sin3x=12

Answer & Explanation

Philip Williams

Philip Williams

Beginner2021-12-27Added 39 answers

Within the range [0,2π]  ,sinπ6=sin5π6=12
So, sin(3x)=sinπ6=sin5π6
Also sine has periodicity of 2π. Thus sin(2nπ+θ)=sinθ where n is an integer.
So,
sin(3x)=sinπ6=sin(2nπ+π6)3x=2nπ+π6
Similarly,
sin(3x)=sin5π6=sin(2nπ+5π6)3x=2nπ+5π6
Jim Hunt

Jim Hunt

Beginner2021-12-28Added 45 answers

When
sinx=sinαx=nπ+(1)nα,
where n=0,±±1,±2,±3, So here
sin3x=sinπ63x=nπ+(1)nπ3x=nπ3+(1)nπ9,   n=0,±1,±3,
nick1337

nick1337

Expert2022-01-08Added 573 answers

sin(x) is periodic, as in it has a period of 2π. Therefore, it will intersect the same y-value every period, or every 2nπ where n is an integer. By scaling θ by 3, you are actually scaling the period of sinθ by 13. Thus the new period is 2π3. You can see this in action with your solution you gave, which was
sinθ=12
θ=sin1(12)
Here's the kicker: sinθ intersects 12 twice every period, so there are two solutions. By inspecting the unit circle, find why this must be true. The two solutions are
3θ=π6
3θ=5π6
But we want to know the solutions for the entire domain (,), so we want
3θ=π6+2nπ
3θ=5π6+2nπ
And now by dividing by 3, you can see why your period is scaled by 13 (as well as the solution):
θ=π18+2nπ3
θ=5π18+2nπ3

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