Question

Sin(x+60)+cos(x-60)=0

Answer 1

Раскрываем синус суммы и косинус разности

$sin(x+60)+cos(x-60)=sinxcos60+sin60cosx+cosxcos60+sinxsin60=\\ \\ =\dfrac{1}{2}sinx+\dfrac{\sqrt{3} }{2} cosx + \dfrac{1}{2} cosx+\dfrac{\sqrt{3} }{2} sinx=\\ \\ = \dfrac{\sqrt{3}+1 }{2} cosx+\dfrac{\sqrt{3}+1 }{2} sinx=0 \\\ \\ cosx+sinx=0$

Делим на cosx отличный от 0

$1+tgx=0\\ \\ tgx=-1\\ \\ x= -\dfrac{\pi }{4} +\pi n$ n целое число

Answer 2

To solve the equation sin(x+60) + cos(x-60) = 0, we can use trigonometric identities and algebraic manipulation.

Step 1: Apply the angle sum identity for sine and cosine:

sin(A + B) = sin(A) * cos(B) + cos(A) * sin(B) cos(A + B) = cos(A) * cos(B) - sin(A) * sin(B)

In our equation, A = x and B = 60:

sin(x + 60) = sin(x) * cos(60) + cos(x) * sin(60) = sin(x) * 1/2 + cos(x) * √3/2 cos(x - 60) = cos(x) * cos(60) - sin(x) * sin(60) = cos(x) * 1/2 - sin(x) * √3/2

Step 2: Substitute these expressions into the original equation:

sin(x+60) + cos(x-60) = 0 (sin(x) * 1/2 + cos(x) * √3/2) + (cos(x) * 1/2 - sin(x) * √3/2) = 0

Step 3: Group the terms with sin(x) and cos(x):

(sin(x) * 1/2 + cos(x) * 1/2) + (cos(x) * √3/2 - sin(x) * √3/2) = 0

Step 4: Combine the like terms:

sin(x) * (1/2 - √3/2) + cos(x) * (1/2 + √3/2) = 0

Step 5: Set each term equal to zero to find the solutions for x:

1/2 - √3/2 = 0 and 1/2 + √3/2 = 0

Solving for x in each equation:

1. 1/2 - √3/2 = 0: 1/2 = √3/2 1 = √3 (Not possible since √3 is an irrational number.)

2. 1/2 + √3/2 = 0: √3/2 = -1/2 √3 = -1 (Not possible since the square root of a positive number cannot be equal to a negative number.)

Conclusion: The given equation sin(x+60) + cos(x-60) = 0 has no real solutions for x. It means there are no values of x that satisfy this equation in the real number domain.