number of real solution of sin ⁡ x ⋅ sin ⁡ 2 x ⋅ sin ⁡ 3 x = 1 for all x ∈ set of real numbers sin ⁡ x ⋅ 2 sin ⁡ x cos ⁡ x ⋅ ( 3 sin ⁡ x − 4 sin 3 ⁡ x ) = 1 2 sin 3 ⁡ x cos ⁡ x ( 3 − 4 sin 2 ⁡ x ) = 1 2 sin 3 ⁡ x ( 3 − 4 sin 2 ⁡ x ) = sec ⁡ xdrawing graph of LHS is hard, i want solution without graphcould some help me with this

Question

number of real solution of   sin  ⁡  x  ⋅  sin  ⁡  2  x  ⋅  sin  ⁡  3  x  =  1 for all   x  ∈ set of real numbers  sin  ⁡  x  ⋅  2  sin  ⁡  x  cos  ⁡  x  ⋅  (  3  sin  ⁡  x  −  4      sin    3    ⁡  x  )  =  1  2      sin    3    ⁡  x  cos  ⁡  x  (  3  −  4      sin    2    ⁡  x  )  =  1  2      sin    3    ⁡  x  (  3  −  4      sin    2    ⁡  x  )  =  sec  ⁡  xdrawing graph of LHS is hard, i want solution without graphcould some help me with this

Gia Schaefer · Accepted Answer

Since       |    sin  ⁡  x      |    ≤  1, this is equivalent to   sin  ⁡  x  =  ±  1 and   sin  ⁡  (  2  x  )  =  ±  1 and   sin  ⁡  (  3  x  )  =  ±  1 with zero or two −1. For zero −1&#039;s, we have  x  =  π      /    2  +  2  π  n  =  π      /    4  +  π  m  ,which is a contradiction. Similar for two −1&#039;s. There are no solutions.

Edith Mayer · Accepted Answer

Hint:Since       |    sin  ⁡  x      |    ≤  1, this can only be true if all of the factors have magnitude 1 and 1 or 3 of them are positive.

number of real solution of sin ⁡ x ⋅ sin &#x206

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