solve 2cos^2x-sinx=1, 0 < or equal to x < 2pie
cos ^ 2 ( x ) = 1 - sin ^ 2 ( x )
2 cos ^ 2 ( x ) - sin ( x ) = 1
2 * [ 1 - sin ^ 2 ( x ) ] - sin ( x ) = 1
2 - 2 sin ^ 2 ( x ) - sin ( x ) = 1 Subtract 1 to both sides
2 - 2 sin ^ 2 ( x ) - sin ( x ) - 1 = 1 - 1
- 2 sin ^ 2 ( x ) - sin ( x ) + 1 = 0
Substitute :
sin ( x ) = u
- 2 u ^ 2 - u + 1 = 0
The exact solution are :
u = 1 / 2
u = - 1
cos ( x ) = 1 / 2
cos ( x ) = - 1
x = 30 ° = pi / 6
x = 270 ° = 9 pi / 6 rad
P.S
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- 2 u ^ 2 - u + 1 = 0
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- 2 u ^ 2 - u + 1 = 0
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Corection:
sin ( x ) = 1 / 2
sin ( x ) = - 1
would the 3 answers be: pi/6, 5pi/6, 3pi/2
Yes
To solve the equation 2cos^2x - sinx = 1 for 0 ≤ x < 2π, we can use algebraic methods to simplify the equation and find the values of x that satisfy it. Here's how you can do it step by step:
Step 1: Rewrite cosine squared as (1 - sin^2x) using the identity cos^2x = 1 - sin^2x:
2(1 - sin^2x) - sinx = 1
Step 2: Distribute the 2:
2 - 2sin^2x - sinx = 1
Step 3: Combine like terms:
-2sin^2x - sinx + 1 = 0
Step 4: Solve the quadratic equation. This equation is quadratic in form, so let's make it easier to solve by substituting u for sinx:
-2u^2 - u + 1 = 0
Step 5: Factor or use the quadratic formula to solve for u. In this case, it does not factor easily, so we will use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / (2a)
Applying the quadratic formula, we have:
u = (-(-1) ± √((-1)^2 - 4(-2)(1))) / (2(-2))
u = (1 ± √(1 + 8)) / (-4)
u = (1 ± √9) / (-4)
u = (1 ± 3) / (-4)
So, we have two possible values for u:
u1 = (1 + 3) / (-4) = -1
u2 = (1 - 3) / (-4) = 1/2
Step 6: Substitute u back in for sinx and solve for x:
For u = -1:
sinx = -1
The only angle that satisfies sinx = -1 for 0 ≤ x < 2π is x = 3π/2.
For u = 1/2:
sinx = 1/2
The two angles that satisfy sinx = 1/2 for 0 ≤ x < 2π are x = π/6 and x = 5π/6.
So, the solution to the equation 2cos^2x - sinx = 1 for 0 ≤ x < 2π is x = 3π/2, π/6, and 5π/6.