What values for Ø (0≤Ø≤2π) satisfy the equation? 2 sin Ø cos Ø + √3 cos Ø=0
We can begin by factoring the equation:
2sinØcosØ + √3cosØ = 0
cosØ(2sinØ + √3) = 0
This equation is satisfied when either:
cosØ = 0 or 2sinØ + √3 = 0
When cosØ = 0, it means that Ø is either π/2 or 3π/2 (since 0 ≤ Ø ≤ 2π).
When 2sinØ + √3 = 0, we can solve for Ø:
2sinØ = -√3
sinØ = -√3/2
This is possible only in the third and fourth quadrants (where sin is negative). Therefore, possible solutions for Ø are:
Ø = 4π/3 or 5π/3
Putting all the solutions together, we get:
Ø = π/2, 3π/2, 4π/3, or 5π/3.
is that correct
Yes, that is correct.
To find the values of Ø that satisfy the equation 2 sin Ø cos Ø + √3 cos Ø = 0, we can first factor out the common term:
cos Ø (2 sin Ø + √3) = 0
Now, we have two possible cases:
Case 1: cos Ø = 0
In this case, Ø could be any angle where cos Ø equals zero. These angles occur at Ø = π/2 and Ø = 3π/2.
Case 2: 2 sin Ø + √3 = 0
To solve this equation for sin Ø, we subtract √3 from both sides:
2 sin Ø = -√3
Dividing by 2, we get:
sin Ø = -√3/2
We know that sin Ø equals -√3/2 at two standard angles: Ø = 5π/3 and Ø = 4π/3.
So, the values for Ø (0≤Ø≤2π) that satisfy the equation are:
Ø = π/2, Ø = 3π/2, Ø = 5π/3, Ø = 4π/3.