Find the values of theta (0,2pi) that satisfy sin theta= square root 3 over 2
Solve 2costheta sintheta-sintheta on (0, 2pi)
please I really need help in knowing how to solve these questions. quite urgent. I don't understand how to do them at all.
You should recognize √3/2 to be a ratio from the 30-60-90° right angled triangle.
e.g. I know sin 60° = √3/2
and 60° = π/3 radians
I also know that the sine is positive in quadrants I and II by the CAST rule,
so Ø = π/3 and π - π/3 or 2π/3 , .......( 60° and 120°)
Ø = π/3 , 2π/3
I assume your second is
2cosØsinØ - sinØ = 0
it factors ...
sinØ(2cosØ - 1) = 0
sinØ = 0 or cosØ = 1/2
Ø = 0,π , 2π
or
Ø = π/3 , 5π/3
Thanks soo much :)
To find the values of θ in the range (0, 2π) that satisfy sin θ = √3/2, we can use the knowledge that sin θ = √3/2 corresponds to the unit circle angles of π/3 and 2π/3. So the solution is θ = π/3 and θ = 2π/3.
Now, let's solve the equation 2cos θsin θ - sin θ on the interval (0, 2π).
Step 1: Factor out sin θ.
sin θ(2cos θ - 1)
Step 2: Set each factor equal to zero and solve.
sin θ = 0 or 2cos θ - 1 = 0
For sin θ = 0, the solutions are θ = 0, π, and 2π.
For 2cos θ - 1 = 0, isolate cos θ:
2cos θ = 1
cos θ = 1/2
This equation corresponds to two angles on the unit circle, which are π/3 and 5π/3.
So, the solutions to 2cos θsin θ - sin θ on the interval (0, 2π) are θ = 0, π, 2π, π/3, and 5π/3.
To find the values of theta (0, 2pi) that satisfy sin(theta) = sqrt(3)/2, we can make use of the unit circle.
Step 1: In the unit circle, the y-coordinate represents the value of sin(theta). Since the y-coordinate is sqrt(3)/2 at two specific angles (2pi/3 and 4pi/3), these are the two values of theta that satisfy sin(theta) = sqrt(3)/2.
Step 2: However, we need to find the values of theta in the range of (0, 2pi). Since the unit circle repeats every 2pi, we can add or subtract multiples of 2pi from our initial solutions to find all the values that satisfy the equation.
Therefore, the values of theta (0, 2pi) that satisfy sin(theta) = sqrt(3)/2 are 2pi/3, 4pi/3, 2pi/3 + 2pi, and 4pi/3 + 2pi. Simplifying those expressions, we get theta = 2pi/3, 4pi/3, 8pi/3, and 10pi/3.
Now, let's move on to solving the equation 2cos(theta)sin(theta) - sin(theta) on the interval (0, 2pi).
Step 1: Notice that the term sin(theta) appears twice. Let's factor out sin(theta) from the equation:
sin(theta) * (2cos(theta) - 1)
Step 2: Now, set each factor equal to zero to find the values of theta that satisfy the equation.
sin(theta) = 0
This occurs at theta = 0, pi, and 2pi.
2cos(theta) - 1 = 0
Solve for cos(theta):
cos(theta) = 1/2
This occurs at theta = pi/3 and 5pi/3.
Therefore, the values of theta (0, 2pi) that satisfy the equation 2cos(theta)sin(theta) - sin(theta) are 0, pi/3, 5pi/3, pi, and 2pi.