Question

Cos7x+sin8x-cos3x+sin2x=0 Решите уравнение.

Answer 1

(cos 7x  -cos3x)  +(sin8x+sin2x) =0;
- 2sin2xsin5x +2sin5xcos3x  = 0;
sin5x(sin2x -cos3x) =0 ;
sin5x=0 ⇒ x =π*k/5 ,k ∈Z .
sin2x -cos3x =0
sin2x -sin(π/2 -3x) =0;
2sin(5x/2 -π/4)*cos(π/4 -x/2) =0 ;
sin(5x/2 -π/4)*cos(x/2 -π/4 ) =0 ;
sin(5x/2 -π/4) =0⇒5x/2  -π/4 =π*k     ⇔        x = π/10+2π/5*k,k ∈Z 
cos(x/2 -π/4)= 0  ⇒x/2 - π/4  =π/2 +π*k ⇔   x=3π/2+2π*k ,k∈Z

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sin2x -cos3x=0;
2sinxcosx -cosx(4cos²x -3) =0;²
cosx(2sinx -4cos²x +3) =0
cosx(4sin²x +2sinx  -1) =0