Ответ: x= π/6 +2*π*n n∈Z
x= 5π/6 +2*π*n n∈Z
x= +-2*π/3 +2*π*n n∈Z
Отбор корней :
x=-11π/6
x=-7π/6
x= -4π/3
Объяснение:
sin(2x) +sin(x) =cos(x) +1/2
sin(2x) = 2*sin(x)*cos(x)
2*sin(x)*cos(x) +sin(x) = cos(x) +1/2
2*sin(x)* (cos(x)+1/2) - (cos(x) +1/2) = 0
(2sin(x)-1)*(cos(x) +1/2) = 0
1) sin(x) = 1/2
x= π/6 +2*π*n n∈Z
x= 5π/6 +2*π*n n∈Z
2) cos(x) = -1/2
x= +-2*π/3 +2*π*n n∈Z
Отбираем корни , для этого необходимо решить неравенства :
1) -5π/2 <= π/6 +2*π*n <= -π </p>
-5/2 <= 1/6 +2n <=-1 ( далее чтобы не писать лишнее буду сокращать на π автоматически)</p>
-15 <= 1+12n <= -5</p>
-16<= 12n <= -6 </p>
-4/3 <= n <= -1/2</p>
Подходит n = -1
x = π/6 -2π = -11π/6
2) -5/2 <= 5/6 +2n <= -1</p>
-15 <= 5+12n <= -6</p>
-20 <=12n<=-11</p>
-5/3 <=n<= -11/12</p>
Подходит n = -1
x= 5π/6 -2πn = -7π/6
3) -5/2 <= +-2/3+2n <= -1</p>
-15 <= +-4+12n <=-6</p>
3.1) -19/12 <=n <= -10/12 (+4)</p>
n=-1
3.2) -11/12 <= n <= -2/12 (-4) </p>
нет целых значений
x= 2π/3 - 2π*n = -4π/3