Find all solutions in the unterval [0,2pi), cos x/2-sinx=0
from sin2A = 2sinAcosA we can say that
sinx = 2sin(x/2)cos(x/2)
cos(x/2) = 2sin(x/2)cos(x/2)
1= 2sin(x/2)
sin(x/2) = 1/2
x/2 = 30°, 150° or π/6, 5π/6 radians
x = 60°, 300° or π/3, 5π/3 radians
To find all solutions to the equation cos(x/2) - sin(x) = 0 in the interval [0, 2π), we can follow these steps:
Step 1: Rewrite the equation to express sine and cosine functions in terms of one trigonometric function.
cos(x/2) - sin(x) = 0
cos(x/2) = sin(x)
cos(x/2) = cos(π/2 - x)
Step 2: Use the identity cos(A) = cos(B) to set the angles equal to each other.
x/2 = π/2 - x
Step 3: Solve the equation for x.
x + 2x/2 = π/2
x + x = π/2
2x = π/2
x = π/4
Step 4: Check if π/4 is in the given interval [0, 2π). It is, so it is a valid solution.
Step 5: Find other solutions by using the periodicity of trigonometric functions.
Since cosine and sine functions are periodic with a period of 2π, we can add or subtract multiples of 2π to the solutions obtained.
x = π/4 + 2πn, where n is an integer.
Step 6: Check which solutions lie in the interval [0, 2π).
Plug in the values of n and check if they satisfy the given interval.
For n = 0: x = π/4
For n = 1: x = π/4 + 2π(1) = 9π/4
For n = 2: x = π/4 + 2π(2) = 17π/4
...
Continue plugging in different values of n to get all the solutions within the given interval.
Hence, the solutions in the interval [0, 2π) are:
x = π/4, 9π/4, 17π/4, ... (where n is an integer)