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Easy

Question

The equation 4 s i n to the power of 2 end exponent invisible function application x minus 2 left parenthesis square root of 3 plus 1 right parenthesis s i n invisible function application x plus square root of 3 equals 0 has

  1. 2 solutions in (0, straight pi)    
  2. 4 solutions in (0,2straight pi)    
  3. 2 Solutions in (-straight pi,straight pi)
  4. 4 Solutions in (-straight pi,straight pi)

The correct answer is: 4 solutions in (0,2straight pi)


    sin to the power of 4 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application xsin invisible function application x plus 2 sin to the power of 2 end exponent invisible function application x plus sin invisible function application x equals 0
    table row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus cos to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x minus 1 plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x plus 1 close square brackets equals 0 end cell row cell s i n invisible function application x open square brackets sin to the power of 3 end exponent invisible function application x plus sin to the power of 2 end exponent invisible function application x plus 2 sin invisible function application x close square brackets equals 0 end cell row cell s i n to the power of 2 end exponent invisible function application x equals 0 text end text text o end text text r end text text end text s i n to the power of 2 end exponent invisible function application x plus s i n invisible function application x plus 2 equals 0 end cell end table
    (not possible for real x)
    text or  end text sin invisible function application x equals 0
    Hence, the solutions are x = 0, p, 2p, 3p.

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