Question

1. √3sinx=2 ; 2. Sinx=π/6 ; 3. √2 cos^2 5x=cos5x ; 4. (2sin2x-cos2x)(1+cos2x)=sin^2 2x ; 5. √3 sinx-tgx+tgx*sinx=√3 ;

Answer 1

√2 cos^2 5x=cos5x
√2cos²5x-cos5x=0
cos5x(2cos5x-1)=0
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5,n∈z
cos5x=1/√2⇒5x=+-π/4+2πk⇒x=+-π/20+2πk/5,k∈z

(2sin2x-cos2x)(1+cos2x)=sin²2x
(2sin2x-cos2x)(1+cos2x)-(1-cos²2x)=0
(2sin2x-cos2x)(1+cos2x)-(1-cos2x)(1+cos2x)=0
(1+cos2x)(2sin2x-cos2x-1+cos2x)=0
(1+cos2x)(2sin2x-1)=0
1+cos2x=0⇒cos2x=-1⇒2x=π+2πn⇒x=π/2+πn,n∈z
2sin2x-1=0⇒sin2x=1/2⇒2x=(-1)^k*π/6+πk⇒x=(-1)^k*π/12+πk/2,k∈z

√3 sinx-tgx+tgx*sinx=√3
√3(sinx-1)+tgx(sinx-1)=0
(sinx-1)(√3+tgx)=0
sinx-1=0⇒sinx=1⇒x=π/2+2πn,n∈z
√3+tgx=0⇒tgx=-√3⇒x=-π/3+πk,k∈z