for sin2x+cosx=0, use double-angle or half-angle formula to simplify the equation and then find all solutions of the equation in the interval (0,2Pie)
Please type your subject in the School Subject box. Any other words, including obscure abbreviations, are likely to delay responses from a teacher who knows that subject well.
sin2x= 2sinxcosx
cosx(2sinx+1)=0
cosx=0 so x=0, PI, 2PI
sinx=-1/2, so x=3PI/2 check that.
To solve the equation sin(2x) + cos(x) = 0 and find all solutions in the interval (0, 2π), we can use the double-angle formula for sine and the half-angle formula for cosine.
First, let's apply the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x)
The equation becomes:
2sin(x)cos(x) + cos(x) = 0
Now, let's rearrange the equation and factor out the common term, cos(x):
cos(x)(2sin(x) + 1) = 0
To find the solutions, we need to consider two cases:
Case 1: cos(x) = 0
In this case, we have cos(x) = 0, which occurs when x = π/2 and x = 3π/2.
Case 2: 2sin(x) + 1 = 0
To solve this equation, we can use the half-angle formula for cosine:
cos(x) = 1 - 2sin^2(x/2)
Substituting this into the equation, we get:
1 - 2sin^2(x/2) = 0
Rearranging, we have:
sin^2(x/2) = 1/2
Taking the square root of both sides, we get:
sin(x/2) = ± √(1/2)
Now, we need to find the values of x/2 that satisfy this equation. Since we are interested in solutions in the interval (0, 2π), we can find the solutions in the interval (0, π) first and then double them for the full solution.
For sin(x/2) = √(1/2), we have:
x/2 = π/4 and x/2 = 3π/4
Solving for x, we find x = π/2 and x = 3π/2.
Similarly, for sin(x/2) = -√(1/2), we have:
x/2 = 5π/4 and x/2 = 7π/4
Solving for x, we find x = 5π/2 and x = 7π/2.
Now, let's compile all the solutions:
x = π/2, 3π/2, π/2 + 2π, 3π/2 + 2π, 5π/2 + 2π, 7π/2 + 2π
Simplifying, we have:
x = π/2, 3π/2, 5π/2, 7π/2, 3π, 7π
Therefore, the solutions of the equation sin(2x) + cos(x) = 0 in the interval (0, 2π) are x = π/2, 3π/2, 5π/2, 7π/2, 3π, and 7π.