1.Solve, finding all solutions in [0, 2π).
cosx sin2x + sinx cosx - sinx = 0
2.Solve, finding all solutions in [0, 2π).
3sec x = 2tan^2x
cosx sin2x + sinx cosx - sinx = 0
cosx (2sinxcosx) + sinxcosx - sinx = 0
sinx(2cos^2x + cosx - 1) = 0
sinx(2cosx-1)(cosx+1) = 0
now it's easy, right?
3secx = 2tan^2x
3secx = 2(sec^2x-1)
2sec^2x - 3secx - 2 = 0
(2secx+1)(secx-2)
...
To solve these equations, we will use trigonometric identities and algebraic manipulation.
1. Solve cos(x)sin(2x) + sin(x)cos(x) - sin(x) = 0 in the interval [0, 2π):
Let's start by simplifying the equation. Rearranging the terms, we have sin(2x)cos(x) + sin(x)cos(x) - sin(x) = 0.
Using the identity sin(2x) = 2sin(x)cos(x), we can substitute this into the equation:
2sin(x)cos(x)cos(x) + sin(x)cos(x) - sin(x) = 0.
2sin(x)cos^2(x) + sin(x)cos(x) - sin(x) = 0.
Factoring out sin(x), we get:
sin(x)(2cos^2(x) + cos(x) - 1) = 0.
Now, we have two cases to consider:
Case 1: sin(x) = 0.
If sin(x) = 0, this implies that x = 0 or x = π.
Case 2: 2cos^2(x) + cos(x) - 1 = 0.
To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring, we get:
(2cos(x) - 1)(cos(x) + 1) = 0.
Setting each factor equal to zero, we have two possibilities:
2cos(x) - 1 = 0, which implies cos(x) = 1/2. Solving for x, we get x = π/3 or x = 5π/3.
cos(x) + 1 = 0, which implies cos(x) = -1. Solving for x, we get x = π.
Therefore, the solutions for the equation cos(x)sin(2x) + sin(x)cos(x) - sin(x) = 0 in the interval [0, 2π) are:
x = 0, x = π, x = π/3, x = 5π/3.
2. Solve 3sec(x) = 2tan^2(x) in the interval [0, 2π):
To solve this equation, we need to rewrite sec(x) and tan^2(x) in terms of sine and cosine.
Recall that sec(x) is the reciprocal of cosine (sec(x) = 1/cos(x)), and tan^2(x) is the square of tangent (tan^2(x) = sin^2(x)/cos^2(x)).
Substituting these expressions into the given equation, we have:
3(1/cos(x)) = 2(sin^2(x)/cos^2(x)).
To eliminate the denominators, we can multiply both sides of the equation by cos^2(x):
3cos^2(x) = 2sin^2(x).
Using the Pythagorean identity sin^2(x) + cos^2(x) = 1, we can rewrite the equation as:
3cos^2(x) = 2(1 - cos^2(x)).
Expanding and rearranging the terms, we get:
3cos^2(x) = 2 - 2cos^2(x).
Combining like terms, we have:
5cos^2(x) = 2.
Dividing by 5, we get:
cos^2(x) = 2/5.
Taking the square root of both sides, we have:
cos(x) = ±√(2/5).
Now, we have two cases to consider:
Case 1: cos(x) = √(2/5).
To find the solutions, we take the inverse cosine (arccos) of both sides, giving us:
x = arccos(√(2/5)) or x = 2π - arccos(√(2/5)).
Case 2: cos(x) = -√(2/5).
Similarly, we take the inverse cosine of both sides, giving us:
x = π - arccos(√(2/5)) or x = π + arccos(√(2/5)).
Therefore, the solutions for the equation 3sec(x) = 2tan^2(x) in the interval [0, 2π) are:
x = arccos(√(2/5)), 2π - arccos(√(2/5)), π - arccos(√(2/5)), π + arccos(√(2/5)).