How to solve the trigonometric equation sin ⁡ x + cos ⁡ x = sin ⁡ 2 x + cos ⁡ 2 x?My attempt: sin ⁡ x + cos ⁡ x = sin ⁡ 2 x + cos ⁡ 2 x ⟹ sin ⁡ x + cos ⁡ x = 2 sin ⁡ x cos ⁡ x + cos 2 ⁡ x − sin 2 ⁡ x ⟹ sin ⁡ x + cos ⁡ x = 2 sin ⁡ x cos ⁡ x + cos 2 ⁡ x − ( 1 − cos 2 ⁡ x ) ⟹ sin ⁡ x + cos ⁡ x = 2 sin ⁡ x cos ⁡ x + 2 cos 2 ⁡ x − 1 ⟹ sin ⁡ x − 2 sin ⁡ x cos ⁡ x + cos ⁡ x − 2 cos 2 ⁡ x = − 1 ⟹ sin ⁡ x ( 1 − 2 cos ⁡ x ) + cos ⁡ x ( 1 − 2 cos ⁡ x ) = − 1 ⟹ ( 1 − 2 cos ⁡ x ) ( sin ⁡ x + cos ⁡ x ) = − 1 ⟹ ( 1 − 2 cos ⁡ x ) = − 1 or ( sin ⁡ x + cos ⁡ x ) = − 1 ⟹ x = 2 n π or sin ⁡ 2 x = 0 ⟹ x = 2 n π or 2 x = n π ∴ x = 2 n π or x = n π 2 But the answers given in my book are x = 2 n π and x = ( 4 n + 1 ) π 6 . Where have I gone wrong? Please help.

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How to solve the trigonometric equation   sin  ⁡  x  +  cos  ⁡  x  =  sin  ⁡  2  x  +  cos  ⁡  2  x?My attempt:  sin  ⁡  x  +  cos  ⁡  x  =  sin  ⁡  2  x  +  cos  ⁡  2  x    ⟹    sin  ⁡  x  +  cos  ⁡  x  =  2  sin  ⁡  x  cos  ⁡  x  +      cos    2    ⁡  x  −      sin    2    ⁡  x    ⟹    sin  ⁡  x  +  cos  ⁡  x  =  2  sin  ⁡  x  cos  ⁡  x  +      cos    2    ⁡  x  −  (  1  −      cos    2    ⁡  x  )    ⟹    sin  ⁡  x  +  cos  ⁡  x  =  2  sin  ⁡  x  cos  ⁡  x  +  2      cos    2    ⁡  x  −  1    ⟹    sin  ⁡  x  −  2  sin  ⁡  x  cos  ⁡  x  +  cos  ⁡  x  −  2      cos    2    ⁡  x  =  −  1    ⟹    sin  ⁡  x  (  1  −  2  cos  ⁡  x  )  +  cos  ⁡  x  (  1  −  2  cos  ⁡  x  )  =  −  1    ⟹    (  1  −  2  cos  ⁡  x  )  (  sin  ⁡  x  +  cos  ⁡  x  )  =  −  1    ⟹    (  1  −  2  cos  ⁡  x  )  =  −  1 or   (  sin  ⁡  x  +  cos  ⁡  x  )  =  −  1    ⟹    x  =  2  n  π or   sin  ⁡  2  x  =  0    ⟹    x  =  2  n  π or   2  x  =  n  π  ∴  x  =  2  n  π or   x  =            n      π        2  But the answers given in my book are   x  =  2  n  π and   x  =            (      4      n      +      1      )      π        6  . Where have I gone wrong? Please help.

Rebecca Benitez · Accepted Answer

(  1  −  2  cos  ⁡  x  )  (  sin  ⁡  x  +  cos  ⁡  x  )  =  −  1    ⟹    (  1  −  2  cos  ⁡  x  )  =  −  1This is an incorrect implication.  a  b  =  c only implies   a  =  c or   b  =  c when   c  =  0For   c  =  −  1 as in this case, you could have   a  =  1  ,  b  =  −  1 or   a  =  5  ,  b  =  −  0.2 or   a  =  −  1000  ,  b  =  0.001 or an infinite number of other combinations.

Widersinnby7 · Accepted Answer

Use Subtraction:  sin  ⁡  2  x  −  sin  ⁡  x  =  cos  ⁡  x  −  cos  ⁡  2  x  2  sin  ⁡      x    2    cos  ⁡            3      x        2    =  2  sin  ⁡            3      x        2    sin  ⁡      x    2  So,  sin  ⁡      x    2    =  0OR  tan  ⁡            3      x        2    =  1

How to solve the trigonometric equation sin x+cos x=sin 2x+cos 2x?

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