Solve: 3 sin ⁡ x − cos ⁡ x = 2, − 2 π &lt; x &lt; 2 πMy attempt: 3 sin ⁡ x − cos ⁡ x = 2 3 2 sin ⁡ x − 1 2 cos ⁡ x = 1 sin ⁡ ( π 3 ) . sin ⁡ x − cos ⁡ ( π 3 ) cos ⁡ x = 1 cos ⁡ ( π 3 ) cos ⁡ x − sin ⁡ ( π 3 ) . sin ⁡ x = − 1 cos ⁡ ( x + π 3 ) = cos ⁡ π x + π 3 = 2 n π ± π , where n belongs to Z x = 2 n π ± π − π 3 Now, if n=0, x = 2 π 3 , − 4 π 3 If n=1, x = 8 π 3 , 2 π 3 If n=-1, x = − 4 π 3 , − 10 π 3 So, in the given interval, x = 2 π 3 , − 4 π 3 Question:When n=0, and when n=1, I get 2 π 3 . Similarly, when n=0, and when n=−1, I get − 4 π 3 . Why am I getting the same angle twice for different values of n?

Question

Solve:       3    sin  ⁡  x  −  cos  ⁡  x  =  2,   −  2  π  &amp;lt;  x  &amp;lt;  2  πMy attempt:      3    sin  ⁡  x  −  cos  ⁡  x  =  2            3        2    sin  ⁡  x  −      1    2    cos  ⁡  x  =  1  sin  ⁡  (      π    3    )  .  sin  ⁡  x  −  cos  ⁡  (      π    3    )  cos  ⁡  x  =  1  cos  ⁡  (      π    3    )  cos  ⁡  x  −  sin  ⁡  (      π    3    )  .  sin  ⁡  x  =  −  1  cos  ⁡  (  x  +      π    3    )  =  cos  ⁡      π    x  +      π    3    =  2  n  π  ±  π  ,      where           n         belongs to                   Z              x  =  2  n  π  ±  π  −      π    3  Now, if n=0,  x  =            2      π        3    ,  −            4      π        3  If n=1,  x  =            8      π        3    ,            2      π        3  If n=-1,  x  =  −            4      π        3    ,  −            10      π        3  So, in the given interval,   x  =            2      π        3    ,            −      4      π        3  Question:When n=0, and when n=1, I get             2      π        3  . Similarly, when n=0, and when n=−1, I get   −            4      π        3  . Why am I getting the same angle twice for different values of n?

Houston Ochoa · Accepted Answer

x  =  2  n  π  ±  π  −      π    3      =  −      π    3    +  π  (      2    n    ±    1    )  .      2    n    ±    1    :    …  ,  −  7  ,  −  7  ,  −  5  ,  −  5  ,  −  3  ,  −  3  ,  −  1  ,  −  1  ,  1  ,  1  ,  3  ,  3  ,  5  ,  5  ,  …  .When solving trigonometric equations generally and the cosine or secant of the generating solution equals −1, then       2    n    ±    1   occurs, giving rise to repeating/overlapping solutions.This is inconsequential since the obtained collection of solutions is exhaustive anyway.

Solve: sqrt(3) sin x−cos x=2, -2 pi<x<2 pi

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