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Verified answer

Task/26315052
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1.
2cos²πx +sinπx-1 =0 ;
2(1-sin²πx)  +sinπx -1 =0 ;
2-2sin²πx +sinπx -1 =0 '
2sin²πx - sinπx -1 =0  ;
sinπx =(1+3)/4 = 1 ⇔ πx =π/2 +2πn , n∈Z . ⇔x =2n +1/2 , n∈Z .
или 
sinπx =(1-3)/4 =-1/2 ⇔πx =(-1)^n*π/6 +πn, n∈Z.⇔x =(1/6)(-1)ⁿ+ n ,n∈Z.
2.
sin⁴x +cos⁴x - cos2x = 0,5 ;
(sin²x +cos²x)² -2sin²x *cos²x - cos2x = 0,5  ;
1 -(1/2sin²2x  - cos2x = 0,5  ;
(1/2)(1-sin²2x)  - cos2x = 0  ;
cos²2x   - 2cos2x =0 ;
cos2x(cos2x- 2) =0 ;
cos2x- 2 =0 ⇔ cos2x = 2 ⇒ x∈∅ .
или
cos2x =0 ⇔ 2x =π/2 +πn , n∈Z  ⇔ x =π/4 +πn/2 , n∈Z. 
3.
cos2x +3sinx +1 =0 ;
1 -2sin²x +3sinx +1 =0 ;
2sin²x -3sinx - 2 =0 ;
sinx =(3+5)/2*2 = 2 ⇒ x∈∅ .
или 
sinx =(3-5)/2*2 = -1/2 ⇔ x =(-1)ⁿ⁺¹  π/6 +πn , n∈Z. 
4.
cos(1,5π-2x) - cosx = 0 ;
- sin2x  - cosx =0 ;
2sinxcosx +cosx =0 ;
2cosx(sinx +1/2) =0 ;
cosx =0  ⇒ x = π/2 +πn , n∈Z 
или 
sinx+1/2 =0 ⇔sinx= -1/2 ⇒ x =(-1)ⁿ⁺¹π/6 +πn , n∈Z. 
5.
cos(2x +π/4) cosx - sin(2x +π/4) sinx = -(√2) / 2 ;
cos(2x +π/4 +x ) = -(√2) / 2 ;
cos(3x +π/4 ) = -(√2) / 2 ;
3x +π/4  = ±(π -π/4) +2πn , n∈Z ;
3x = ±3π/4 -π/4 +2πn , n∈Z.
x =(π/3)*(2n -1),   n ∈ Z 
или 
x = (π/6)*(4n+1) , n∈Z.