How does one solve sin⁡x−3cos⁡x=1 I thought this one up, but I am not sure how to solve it. Here is my attempt: sin⁡x−3cos⁡x=1 (sin⁡x−3cos⁡x)2=1 sin2x−23sin⁡xcos⁡x+3cos2x=1 1−23sin⁡xcos⁡x+2cos2x=1 2cos2x−23sin⁡xcos⁡x=0 2cos⁡x(cos⁡x−3sin⁡x)=0 2cos⁡x=0⇒x∈{π2(2n−1):n∈Z} But how do I solve cos⁡x−3sin⁡x=0

Name: How does one solve sin⁡x−3cos⁡x=1 I thought this one up, but I am not sure how to solve it. Here is my attempt: sin⁡x−3cos⁡x=1 (sin⁡x−3cos⁡x)2=1 sin2x−23sin⁡xcos⁡x+3cos2x=1 1−23sin⁡xcos⁡x+2cos2x=1 2cos2x−23sin⁡xcos⁡x=0 2cos⁡x(cos⁡x−3sin⁡x)=0 2cos⁡x=0⇒x∈{π2(2n−1):n∈Z} But how do I solve cos⁡x−3sin⁡x=0
Uploaded: 2022-02-23T17:22:57+00:00
Description: How does one solve sin⁡x−3cos⁡x=1 I thought this one up, but I am not sure how to solve it. Here is my attempt: sin⁡x−3cos⁡x=1 (sin⁡x−3cos⁡x)2=1 sin2x−23sin⁡xcos⁡x+3cos2x=1 1−23sin⁡xcos⁡x+2cos2x=1 2cos2x−23sin⁡xcos⁡x=0 2cos⁡x(cos⁡x−3sin⁡x)=0 2cos⁡x=0⇒x∈{π2(2n−1):n∈Z} But how do I solve cos⁡x−3sin⁡x=0

Question

How does one solve sin⁡x−3cos⁡x=1  I thought this one up, but I am not sure how to solve it. Here is my attempt:  sin⁡x−3cos⁡x=1   (sin⁡x−3cos⁡x)2=1   sin2x−23sin⁡xcos⁡x+3cos2x=1   1−23sin⁡xcos⁡x+2cos2x=1   2cos2x−23sin⁡xcos⁡x=0   2cos⁡x(cos⁡x−3sin⁡x)=0   2cos⁡x=0⇒x∈{π2(2n−1):n∈Z}  But how do I solve  cos⁡x−3sin⁡x=0

chukizosv · Accepted Answer

Hint: at the very beginning divide both sides by 2 and use the formula for the sin of difference of 2 arguments

vasselefa · Accepted Answer

Hint :  cos⁡x−3sin⁡x=0⇔sin⁡xcos⁡x=33⇔tan⁡x=33  Note : You can divide by cos⁡x since if the case was cos⁡x=0 it would be sin⁡x=±1 and thus the equation would yield ±3±0, thus no problems in the final solution, as the cos zeros are no part of it.