hey, i would really appreciate some help solving for x when:
sin2x=cosx
Use the identity sin 2A = 2sinAcosA
so:
sin 2x = cos x
2sinxcosx - cosx = 0
cosx(2sinx - 1)=0
cosx = 0 or 2sinx=1, yielding sinx=1/2
from cosx=0 and by looking at the cosine graph, we conclude that x=pi/2 or 3pi/2 (90º and 270º)
from sinx = 1/2 and knowing that the sine is positive in the first and second quadrants, we conclude that x = pi/6 or 5pi/6 (30º and 150º)
those are the solutions in the domain 0 ≤ x ≤ 2pi,
if you want a general solution, add 2(k)pi to each of the four solutions I gave you, where k is an integer.
To solve the equation sin 2x = cos x, we can use the identity sin 2A = 2 sin A cos A. This allows us to rewrite the equation as 2 sin x cos x - cos x = 0.
Factoring out the common factor of cos x, we have cos x (2 sin x - 1) = 0.
Now, we can set each factor equal to zero and solve for x:
1) cos x = 0:
We know that the cosine function is zero at π/2 (90°) and 3π/2 (270°) in the given domain (0 ≤ x ≤ 2π).
2) 2 sin x - 1 = 0:
Solving this equation, we get sin x = 1/2. In the first and second quadrants, the sine function is positive, and it is equal to 1/2 at π/6 (30°) and 5π/6 (150°) within the given domain.
Thus, the solutions for 0 ≤ x ≤ 2π are:
x = π/2, 3π/2, π/6, 5π/6.
If you want a general solution, you can express it by adding 2πk to each of the four solutions, where k is an integer.
To solve for x when sin2x = cosx, let's go through the steps:
Step 1: Use the identity sin 2A = 2sinAcosA to rewrite sin2x as 2sinxcosx:
2sinxcosx - cosx = 0
Step 2: Factor out the common factor cosx:
cosx(2sinx - 1) = 0
Step 3: Set each factor equal to zero and solve for x:
cosx = 0 or 2sinx = 1
Step 4: Solve for x when cosx = 0:
By looking at the cosine graph, we can see that x = pi/2 or 3pi/2 (90º and 270º) when cosx = 0.
Step 5: Solve for x when 2sinx = 1:
Using the property of sine being positive in the first and second quadrants, we can conclude that x = pi/6 or 5pi/6 (30º and 150º) when sinx = 1/2.
These are the solutions in the domain 0 ≤ x ≤ 2pi. If you want a general solution, you can add 2(k)pi to each of the four solutions, where k is an integer.