Finding solution of a Trigonometric equationtan⁡A+tan⁡2A+tan⁡3A=0I tried converting these all in sin and cos and I got the answer but the answer didn't match

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Finding solution of a Trigonometric equationtan⁡A+tan⁡2A+tan⁡3A=0I tried converting these all in sin and cos and I got the answer but the answer didn&#039;t match

Hattie Schaeffer · Accepted Answer

Use this:  tan⁡3A=tan⁡A+tan⁡2A1−tan⁡Atan⁡2A  Substitute the numerator with −tan⁡3A and youre

enhebrevz · Accepted Answer

tan⁡A+tan⁡2A=sin⁡3Acos⁡Acos⁡2A tan⁡A+tan⁡2A+tan⁡3A=0⇒sin⁡3Acos⁡Acos⁡2A+sin⁡3Acos⁡3A=0 ⇔sin⁡3A(cos⁡3A+cos⁡Acos⁡2A)=0 If sin⁡3A=0,  3A=nπ where n is any integer Otherwise, cos⁡3A+cos⁡Acos⁡2A=0⇔4cos⁡3A−3cos⁡A+cos⁡A(2cos2A−1)=0 ⇔cos⁡A(6cos2A−4)=0 If cos⁡A=0,   A=(2m+1)π2 Otherwise, 6cos2A−4=0⇔cos2A=23,  cos⁡2A=2cos2A−1=13

nick1337 · Accepted Answer

Hint The following identities for the tangent of multiple angles should be useful tan⁡2A=2tan⁡A1−2tan2⁡A tan⁡3A=3tan⁡A−tan3⁡A1−3tan2⁡A I am sure that you can take from here.

Finding solution of a Trigonometric equation \tan A+\tan 2A+\tan 3A=0 I tried

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