Question

Sinx*tgx=1/2

Answer 1

sin(x)*tg(x)=1/2

sin(x)*sin(x)/cos(x)=1/2

sin^2(x)/cos(x)=1/2

sin^2(x)=(1/2)*cos(x)

1-cos^2(x)=(1/2)*cos(x)

2-2cos^2(x)=cos(x)

2cos^2(x)+cos(x)-2=0

cos(x)=t

2t^2+t-2=0

D=b^2-4ac=1+16=17

t1,2=(-b±sqrt(D)/2a

t1=(-1-sqrt(17))/4

t2=(-1+sqrt(17)/4

То есть

 

1. cos(x)= (-1-sqrt(17))/4<-1 не удовлетворяет ОДЗ

 

2. cos(x)= (-1+sqrt(17))/4

x=±arcos(-1+sqrt(17))/4)+2*pi*n

Ответ:

x=±arcos(-1+sqrt(17))/4)+2*pi*n

Answer 2

To solve the equation sin(x)*tan(x) = 1/2, we can use the identity:

tan(x) = sin(x)/cos(x)

Substituting this identity into the equation, we get:

sin(x)*(sin(x)/cos(x)) = 1/2

Multiplying both sides by cos(x), we get:

sin^2(x) = (1/2)*cos(x)

Using the identity sin^2(x) + cos^2(x) = 1, we can substitute cos^2(x) = 1 - sin^2(x) and get:

sin^2(x) = (1/2)*(1 - sin^2(x))

Simplifying the equation, we get:

2*sin^2(x) = 1 - sin^2(x)

Adding sin^2(x) to both sides, we get:

3*sin^2(x) = 1

Dividing both sides by 3, we get:

sin^2(x) = 1/3

Taking the square root of both sides, we get:

sin(x) = ±sqrt(1/3)

Since the sine function is positive in the first and second quadrants, we have:

sin(x) = sqrt(1/3)

Using the identity tan(x) = sin(x)/cos(x), we can find the value of tan(x):

tan(x) = sin(x)/cos(x) = (sqrt(1/3))/sqrt(2/3) = sqrt(1/2)

Therefore, the solution to the equation sin(x)*tan(x) = 1/2 is:

x = arcsin(sqrt(1/3)) + kpi, where k is an integer or x = pi - arcsin(sqrt(1/3)) + kpi, where k is an integer

(Note: arcsin denotes the inverse sine function, which returns the angle whose sine is the given value. The values of arcsin(sqrt(1/3)) and pi - arcsin(sqrt(1/3)) can be approximated as 0.615 and 2.527 radians, respectively, or approximately 35.264 and 144.736 degrees, respectively.)