3sinx+7cosx=-1 x=

To solve the equation 3sin(x) + 7cos(x) = -1, we can use the trigonometric identity: sin^2(x) + cos^2(x) = 1.

First, let's square the equation to eliminate the sine term: (3sin(x))^2 + 2(3sin(x))(7cos(x)) + (7cos(x))^2 = (-1)^2 9sin^2(x) + 42sin(x)cos(x) + 49cos^2(x) = 1

Using the identity sin^2(x) + cos^2(x) = 1, we can replace sin^2(x) with 1 - cos^2(x): 9(1-cos^2(x)) + 42sin(x)cos(x) + 49cos^2(x) = 1 9 - 9cos^2(x) + 42sin(x)cos(x) + 49cos^2(x) = 1 58cos^2(x) + 42sin(x)cos(x) - 8 = 0

Now, let's use the double angle formula for cosine: cos(2x) = 2cos^2(x) - 1 Substituting this into the equation, we get: 58(1/2)(1 + cos(2x)) + 42sin(x)cos(x) - 8 = 0 29(1 + cos(2x)) + 42sin(x)cos(x) - 8 = 0 29 + 29cos(2x) + 42sin(x)cos(x) - 8 = 0 21cos(2x) + 42sin(x)cos(x) + 21 = 0 cos(2x) + 2sin(x)cos(x) + 1 = 0 cos(2x) + sin(2x) + 1 = 0

Now, let's use the double angle formula for sine: sin(2x) = 2sin(x)cos(x) Substituting this into the equation, we get: cos(2x) + 2(2sin(x)cos(x)) + 1 = 0 cos(2x) + 4sin(x)cos(x) + 1 = 0

Let's substitute u = cos(x): cos(2x) + 4sin(x)cos(x) + 1 = 0 cos(2x) + 2sin(2x) + 1 = 0

Now, we have a quadratic equation in terms of cos(2x): cos^2(2x) + 2sin(2x) + 1 = 0

Using the Pythagorean identity: sin^2(2x) + cos^2(2x) = 1, we can replace sin^2(2x) with 1 - cos^2(2x): (1 - cos^2(2x)) + cos(2x) + 1 = 0 2 - cos^2(2x) + cos(2x) = 0 cos^2(2x) - cos(2x) - 2 = 0

Now, let's solve this quadratic equation for cos(2x): (cos(2x) - 2)(cos(2x) + 1) = 0

Setting each factor equal to zero, we have two possible solutions: cos(2x) - 2 = 0 or cos(2x) + 1 = 0

For cos(2x) - 2 = 0: cos(2x) = 2 Since the range of cosine is -1 to 1, there are no solutions for this equation.

For cos(2x) + 1 = 0: cos(2x) = -1 Using the inverse cosine function (arccos), we can find the values of 2x: 2x = arccos(-1) 2x = π x = π/2 + kπ, where k is an integer.

Therefore, the equation 3sin(x) + 7cos(x) = -1 has infinitely many solutions given by x = π/2 + kπ, where k is an integer