sin3x/sinx-sinx/sin3x=2cos2x

Ответы 1

$\frac{sin3x}{sinx} - \frac{sinx}{sin3x}=2cos2x \\ \\ \frac{sin^23x-sin^2x}{sinx*sin3x}=2cos2x \\ \\ \frac{(sin3x-sinx)(sin3x+sinx)}{sinx*sin3x}=2cos2x\\ \\ \frac{2sinx*cos2x*2sin2x*cosx}{sinx*sin3x}=2cos2x\\ \\ \left \{ {{\frac{2cos2x*sin2x*cosx}{sin3x}=cos2x} \atop {sinx eq 0}} ight. \\ \\ \left \{ {{\frac{cos2x*(2sin2x*cosx-sin3x)}{sin3x}=0} \atop {sinx eq 0}} ight. \\ \\ \left \{ {{\frac{cos2x*(2sin2x*cosx-sin(2x+x))}{sin3x}=0} \atop {sinx eq 0}} ight.$ $\left \{ {{\frac{cos2x*(2sin2x*cosx-(sin2x*cosx+cos2x*sinx))}{sin3x}=0} \atop {sinx eq 0}} ight. \\ \\ \left \{ {{\frac{cos2x*(sin2x*cosx-cos2x*sinx)}{sin3x}=0} \atop {sinx eq 0}} ight. \\ \\ \left \{ {\frac{cos2x*sin(2x-x)}{sin3x}=0} }\atop {sinx eq 0}} ight. \\ \\ \left \{ {{\frac{cos2x*sinx}{sin3x}=0} \atop {sinx eq 0}} ight. \\ \\ \{ {{\frac{(1-2sin^2x)*sinx}{sin3x}=0} \atop {sinx eq 0}} ight.$ $\{ {{sin^2x= \frac{1}{2}} \atop {sinx eq 0,sin3x eq 0}} ight. \\ \\ \{ {{sinx= (+/-)\frac{ \sqrt{2} }{2}} \atop {sinx eq 0,sin3x eq 0}} ight. \\ \\ \{ {{x= \frac{ \pi }{4}+ \frac{ \pi n}{2} ,nEZ} \atop {x eq \frac{ \pi k}{3},kEZ }} ight.$ x = π/4 + πn/2, n ∈ Z
- Автор:
  harryntaf
- 5 лет назад
- 0
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