-2sin(n/9-x/9)= 2sin(n/12-x/12)=-1 2sin^2-3sinx-2=0

Предыдущий
вопрос Следующий
вопрос

The equation given is in trigonometric form and involves the sine function.

First, we can simplify the equation
-2sin(n/9 - x/9) = 2sin(n/12 - x/12)

Dividing both sides by 2 gives
-sin(n/9 - x/9) = sin(n/12 - x/12)

Then, we can use the trigonometric identity sin(a - b) = sin(a)cos(b) - cos(a)sin(b) to simplify further
-sin(n/9)cos(x/9) + cos(n/9)sin(x/9) = sin(n/12)cos(x/12) - cos(n/12)sin(x/12)

Next, we can simplify the equation by expanding the trigonometric functions
-(sin(n)cos(x) - cos(n)sin(x)) = sin(n/12)cos(x/12) - cos(n/12)sin(x/12)

Expanding further gives
-sin(n)cos(x) + cos(n)sin(x) = sin(n)cos(x) - cos(n)sin(x)

Since these two equations are equal, we can see that the equation simplifies to
cos(n)sin(x) = cos(n)sin(x)

This equation holds true for all x and n, as it is an identity.

Regarding the second part of the question, the equation 2sin^2 - 3sin(x) - 2 = 0 is a quadratic equation in terms of sin(x). We can solve it using the quadratic formula
sin(x) = [3 ± sqrt((3^2) - 4(2)(-2))] / (2(2))

sin(x) = [3 ± sqrt(9 + 16)] /
sin(x) = [3 ± sqrt(25)] /
sin(x) = [3 ± 5] / 4

This gives two possible solutions for sin(x):

Since the sine function is bound between -1 and 1, the second solution sin(x) = -1/2 is the valid solution.