In the interval
0 'less than or equal to' x 'less than or equal to' pi
find the values that satisfy the equation
cos 2x - sin x = 0
why not write 0 <= x <= pi
cos2x - sinx = 0
1 - 2sin^2 x - sinx = 0
(2sinx-1)(sinx+1) = 0
not so hard now, eh?
The answer options I get are either
a. 0, pi/3
b. 0, pi/6, pi/3, pi/2
which one would you say suits best?
sinx = 1/2 at x = pi/6, 5pi/6
sinx = -1 at x = 3pi/2, but that's not in the domain.
where did you come up with 0,pi/3,pi/2?
so, pi/6, 5pi/6 are the only solutions. See
http://www.wolframalpha.com/input/?i=cos2x+-+sinx+%3D+0
To find the values of x that satisfy the equation cos 2x - sin x = 0 in the given interval 0 ≤ x ≤ π, we can follow these steps:
Step 1: Rewrite the equation using double-angle and half-angle identities.
Using the double-angle identity for cosine and the identity for sine squared, we can rewrite the equation as:
cos² x - sin x = 0
Step 2: Substitute cos² x using the Pythagorean identity.
Substituting cos² x with 1 - sin² x (using the Pythagorean identity for cosine), we get:
1 - sin² x - sin x = 0
Step 3: Rearrange the equation.
Rearrange the equation to bring all terms to one side:
-sin² x - sin x + 1 = 0
Step 4: Factor the equation.
The equation can be factored as:
(-sin x + 1)(sin x + 1) = 0
Now, we have two factors:
1) -sin x + 1 = 0
2) sin x + 1 = 0
Step 5: Solve for x.
For the first factor, -sin x + 1 = 0, solve for x:
-sin x + 1 = 0
sin x = 1
x = π/2
For the second factor, sin x + 1 = 0, solve for x:
sin x = -1
x = 3π/2 (or 270°)
Step 6: Check the solutions.
Check if the solutions, x = π/2 and x = 3π/2, lie within the given interval 0 ≤ x ≤ π.
- For x = π/2:
0 ≤ π/2 ≤ π - This solution is valid.
- For x = 3π/2:
0 ≤ 3π/2 ≤ π - This solution is not valid because 3π/2 is outside the given interval.
Therefore, the only value that satisfies the equation cos 2x - sin x = 0 in the interval 0 ≤ x ≤ π is x = π/2.