Помогите решить

12. Найти сумму корней на промежутке (ctgx+√3) sin2x=0, на [-100˚; 300˚]

13. (√3ctgx-1) cos (5pi/2-x) = 0, на [-150°; 250°]

14. Cos (pi/2+2x) = √2sin (pi+x), на[-100˚; 120˚] тоже найти сумму корней

На промежутке

12)

ctgx=-V3⇒x=5π/6+πn, n∈z

sin2x=0⇒2x=πk⇒x=πk/2, k∈z

x={5π/6; 0; π}

5π/6+0+π=11π/6

13)

ctgx=1/V3⇒x=π/3+πn, n∈z

cos (5π/2-x) = sinx

sinx=0⇒x=πk, k∈z

x={π/3; 4π/3; 0; π}

π/3+4π/3+0+π=8π/3

14)

cos (π/2+2x) = - sin2x

sin (π+x) = - sinx

sin2x-V2sinx=0

sinx (2cosx-V2) = 0

sinx=0⇒x=πn. n∈z

cosx=V2/2⇒x=+-π/4+2πk, k∈z

x={0; -π/4; π/4}

0-π/4+π/4=0

(V-Корень)

12

ctgx=-√3⇒x=5π/6+πn, n∈z

sin2x=0⇒2x=πk⇒x=πk/2, k∈z

x={5π/6; 0; π}

5π/6+0+π=11π/6

13

ctgx=1/√3⇒x=π/3+πn, n∈z

cos (5π/2-x) = sinx

sinx=0⇒x=πk, k∈z

x={π/3; 4π/3; 0; π}

π/3+4π/3+0+π=8π/3

14

cos (π/2+2x) = - sin2x

sin (π+x) = - sinx

sin2x-√2sinx=0

sinx (2cosx-√2) = 0

sinx=0⇒x=πn. n∈z

cosx=√2/2⇒x=+-π/4+2πk, k∈z

x={0; -π/4; π/4}

0-π/4+π/4=0