Approximate the equation's soultions in the interval (o, 2pi). If possible find the exact solutions.
sin 2x sinx = cosx
I do not know where to start.
To approximate the solution and find the exact solutions for the given equation, sin(2x)sin(x) = cos(x) in the interval (0, 2π), you can follow these steps:
1. Rearrange the equation to have zero on one side:
sin(2x)sin(x) - cos(x) = 0
2. Use trigonometric identities to simplify the equation:
Rewrite sin(2x) in terms of sin(x) and cos(x) using the double-angle identity:
2sin(x)cos(x)sin(x) - cos(x) = 0
3. Factor out the common term cos(x) from the equation:
cos(x)(2sin(x)sin(x) - 1) = 0
4. Set each factor equal to zero and solve:
a) cos(x) = 0
In the interval (0, 2π), cos(x) equals zero at π/2 and 3π/2.
b) 2sin(x)sin(x) - 1 = 0
To solve this equation, use the identity sin^2(x) + cos^2(x) = 1:
Rewrite the equation as 2sin^2(x) - 1 = 0
2sin^2(x) = 1
sin^2(x) = 1/2
Take the square root of both sides:
sin(x) = ±√(1/2)
sin(x) = ±1/√2
sin(x) = ±(1/√2)
sin(x) = ±(√2/2)
In the interval (0, 2π), sin(x) equals √2/2 at π/4 and 3π/4.
Similarly, sin(x) equals -√2/2 at 5π/4 and 7π/4.
5. Combine all the solutions obtained earlier:
The exact solutions in the interval (0, 2π) are:
x = π/2, 3π/2, π/4, 3π/4, 5π/4, 7π/4
To approximate the solutions, you can use numerical methods such as the bisection method or Newton's method, or use a graphing calculator to find the x-values where the equation is satisfied.