Question

1)sin6x+sin4x=cosx 2)cosx-√3sinx=√2 3)cos2x=cosx*cos3x помогите решить пожалуйста,срочно!!!))

Answer 1

2) cosx-√3sinx=√2 |:2
½cosx-√3/2sinx=√2/2
½=sin(π/6); √3/2=cos(π/6)
sin(π/6)cosx-cos(π/6)sinx=√2/2
sin(x-π/6)=-√2/2
x-π/6=(-1)^n•arcsin(-√2/2)+ πn
x-π/6=(-1)^(n+1)•arcsin(√2/2)+ πn
x=(-1)^(n+1)•π/4+π/6+ πn

Answer 2

Equation 1: sin(6x) + sin(4x) = cos(x)

To solve the equation sin(6x) + sin(4x) = cos(x), we can use trigonometric identities and algebraic manipulation.

First, let's rewrite the equation using trigonometric identities: sin(6x) + sin(4x) = cos(x) 2sin(5x)cos(x) = cos(x)

Now, we can simplify the equation by canceling out the common factor of cos(x): 2sin(5x) = 1

To isolate sin(5x), divide both sides of the equation by 2: sin(5x) = 1/2

To find the values of x that satisfy this equation, we can use the inverse sine function (sin^(-1)): 5x = sin^(-1)(1/2)

Using a calculator, we can find the principal value of sin^(-1)(1/2) to be π/6 or approximately 0.5236 radians.

Now, we can solve for x by dividing both sides of the equation by 5: x = (π/6) / 5

Therefore, the solution to the equation sin(6x) + sin(4x) = cos(x) is x = (π/6) / 5.

Equation 2: cos(x) - √3sin(x) = √2

To solve the equation cos(x) - √3sin(x) = √2, we can again use trigonometric identities and algebraic manipulation.

First, let's rewrite the equation using trigonometric identities: cos(x) - √3sin(x) = √2

Now, we can rearrange the equation to isolate sin(x): √3sin(x) = cos(x) - √2

Next, we can square both sides of the equation to eliminate the square root: 3sin^2(x) = cos^2(x) - 2√2cos(x) + 2

Using the identity sin^2(x) + cos^2(x) = 1, we can substitute it into the equation: 3(1 - cos^2(x)) = cos^2(x) - 2√2cos(x) + 2

Expanding and rearranging the equation, we get: 3cos^2(x) - 2√2cos(x) - 1 = 0

Now, we have a quadratic equation in terms of cos(x). We can solve this equation using the quadratic formula: cos(x) = [2√2 ± √(8 + 12)] / (2 * 3)

Simplifying the equation further, we get: cos(x) = (√2 ± √5) / 3

Using a calculator, we can find the values of cos(x) to be approximately 0.934 and -0.401.

To find the corresponding values of x, we can use the inverse cosine function (cos^(-1)): x = cos^(-1)((√2 ± √5) / 3)

Therefore, the solutions to the equation cos(x) - √3sin(x) = √2 are x = cos^(-1)((√2 + √5) / 3) and x = cos^(-1)((√2 - √5) / 3).

Equation 3: cos(2x) = cos(x) * cos(3x)

To solve the equation cos(2x) = cos(x) * cos(3x), we can again use trigonometric identities and algebraic manipulation.

First, let's rewrite the equation using trigonometric identities: cos(2x) = cos(x) * cos(3x)

Now, we can use the double-angle identity for cosine: 2cos^2(x) - 1 = cos(x) * (4cos^3(x) - 3cos(x))

Expanding and rearranging the equation, we get: 2cos^2(x) - 1 = 4cos^4(x) - 3cos^2(x)

Combining like terms, we have: 4cos^4(x) - 5cos^2(x) + 1 = 0

Now, we have a quadratic equation in terms of cos^2(x). We can solve this equation by factoring: (2cos^2(x) - 1)(2cos^2(x) - 1) = 0

Taking the square root of both sides, we get: 2cos^2(x) - 1 = 0

Solving for cos(x), we have: cos(x) = ±√(1/2)

Using a calculator, we can find the values of cos(x) to be approximately ±0.707.

To find the corresponding values of x, we can use the inverse cosine function (cos^(-1)): x = cos^(-1)(±√(1/2))

Therefore, the solutions to the equation cos(2x) = cos(x) * cos(3x) are x = cos^(-1)(√(1/2)) and x = cos^(-1)(-√(1/2)).

Please note that these solutions are approximate values, and there may be other solutions depending on the range of x specified.