Solve each equation, giving all solutions (general solutions), tan5(3x)=9tan⁡(3x)

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nitruraviX · Accepted Answer

Let&#039;s make it simple and let y=tan⁡(3x), then y5=9y, so y(y4−9)=y(y2−3)(y2+3)=0. So y=0,±3 are the real roots. Now we go back to y=tan⁡(3x)=0,±3. So x=0,tan⁡(3x)=±3.tan⁡(60)=3, so 3x=60andx=20 degrees. But we know that tangent is negative in QII and QIV. So 3x=180−60=120andx=40 degrees, 3x=360−60=300 so x=100 degrees. CHECK: x=0: tan⁡(0)=0 so tan5(3x)=tan⁡(3x)=0. x=20: 3x=60, tan⁡(60)=3andtan5(60)=93=9tan⁡(60). x=40: 3x=120, tan⁡(120)=−3andtan5(120)=−93=9tan⁡(120). x=100: 3x=300, tan⁡(300)=−3andtan5(300)=−93=9tan⁡(300). But we can add 360 to 0, 60, 120 and 300, then divide by 3 to find all possible angles, so the solution is: x=(13)(360n+0)=120n,(13)(360n+60)=120n+20,120n+40,120n+100 where n is an integer.

Solve each equation, giving all solutions (general solutions), tan^5(3x)=9tan(3x)

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