I'm stuck at a relatively simple equation where I was simplifying like this: cos 2 ⁡ ( x ) + 2 sin ⁡ ( x ) cos ⁡ ( x ) − sin 2 ⁡ ( x ) = 1 ⟺ sin ⁡ ( 2 x ) + cos ⁡ ( 2 x ) = 1How to I continue?

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I&#039;m stuck at a relatively simple equation where I was simplifying like this:      cos    2    ⁡  (  x  )  +  2  sin  ⁡  (  x  )  cos  ⁡  (  x  )  −      sin    2    ⁡  (  x  )  =  1    ⟺    sin  ⁡  (  2  x  )  +  cos  ⁡  (  2  x  )  =  1How to I continue?

lavarcar2d2 · Accepted Answer

Rewrite  cos  ⁡  (  x      )    2    +  2  sin  ⁡  (  x  )  cos  ⁡  (  x  )  −  sin  ⁡  (  x      )    2    =  1use the trigonometric identity   sin  ⁡  (  x      )    2    +  cos  ⁡  (  x      )    2    =  1 to get  cos  ⁡  (  x      )    2    +  2  sin  ⁡  (  x  )  cos  ⁡  (  x  )  −  sin  ⁡  (  x      )    2    =  cos  ⁡  (  x      )    2    +  sin  ⁡  (  x      )    2    .Now, simplify to get  2  sin  ⁡  (  x  )  cos  ⁡  (  x  )  −  2  sin  ⁡  (  x      )    2    =  0.Factoring the LHS gives  2  sin  ⁡  (  x  )  (  cos  ⁡  (  x  )  −  sin  ⁡  (  x  )  )  =  0.Now, all you need to solve is   sin  ⁡  (  x  )  =  0 and   cos  ⁡  (  x  )  =  sin  ⁡  (  x  )There is no perfect system for solving these types of equations. In general, avoid things like   arcsin  ⁡  (  x  ) or   sin  ⁡  (  x  )  =  ±      1    −    cos    ⁡    (    x          )      2       before you&#039;ve exhausted the other options. Also, always try to replace 1 by   sin  ⁡  (  x      )    2    +  cos  ⁡  (  x      )    2   just to see if that might work.

Simplify like this: cos^2(x)+2sin(x)cos(x)−sin^2(x)=1 <=> sin(2x)+cos(2x)=1 <=> cos(2x)+cos((pi)/2+2x)=1 How to I continue?

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