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

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Philip Williams · Accepted Answer

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

Jim Hunt · Accepted Answer

When   sin⁡x=sin⁡α⇒x=nπ+(−1)nα,   where n=0,±±1,±2,±3,… So here  sin⁡3x=sin⁡π6⇒3x=nπ+(−1)nπ3⇒x=nπ3+(−1)nπ9,   n=0,±1,±3,…

nick1337 · Accepted Answer

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 sin⁡θ by 13. Thus the new period is 2π3. You can see this in action with your solution you gave, which was sin⁡θ=12 θ=sin−1⁡(12) Here&#039;s the kicker: sin⁡θ 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

If I have a such equation to solve:\sin 3x=\frac12

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