2(sinx)^2-5cosx-4=0
Solve for all solutions between [0,2pi]
change sin^2x to 1-cos^2x, then
2-2cos^2x-5cosx-4=0
let u= cosx
2-2u^2-5u-4=0
2u^2+5u+2=0
(2u+1)(U+2)=0
u=-1/2 u=-2
cosx=-1/2
x=-2PI/3, 2/3 PI
between o, 2PI then x= 2/3 PI, and 4/3 PI
put those two angles into the original equation, and check.
To solve the equation 2(sinx)^2 - 5cosx - 4 = 0 for all solutions between [0, 2π], you can follow these steps:
Step 1: Rearrange the equation
Move the constant term to the other side of the equation:
2(sinx)^2 - 5cosx = 4
Step 2: Use the trigonometric identity
Remember the identity sin^2(x) + cos^2(x) = 1. Rearrange it to solve for cos^2(x):
cos^2(x) = 1 - sin^2(x)
Step 3: Substitute the identity into the equation
Replace cos^2(x) with 1 - sin^2(x) in the original equation:
2(sinx)^2 - 5(1 - sin^2(x)) = 4
Simplify the equation:
2(sinx)^2 - 5 + 5sin^2(x) = 4
2(sinx)^2 + 5sin^2(x) = 9
Step 4: Combine like terms
Combine the terms on the left side of the equation:
7sin^2(x) + 2(sinx)^2 = 9
Step 5: Substitute sin^2(x) with 1 - cos^2(x)
Use the identity sin^2(x) + cos^2(x) = 1 to replace sin^2(x) in the equation:
7(1 - cos^2(x)) + 2(sinx)^2 = 9
7 - 7cos^2(x) + 2(sinx)^2 = 9
Simplify the equation:
7cos^2(x) - 2(sinx)^2 = -2
Step 6: Use the trigonometric identity
Rearrange the identity cos(2x) = cos^2(x) - sin^2(x) to solve for cos^2(x) - sin^2(x):
cos^2(x) - sin^2(x) = cos(2x)
Replace cos^2(x) - sin^2(x) with cos(2x) in the equation:
7cos(2x) - 2(sinx)^2 = -2
Step 7: Use the Pythagorean identity
Substitute 1 - cos^2(2x) for sin^2(2x) using the Pythagorean identity sin^2(2x) + cos^2(2x) = 1:
7cos(2x) - 2(1 - cos^2(2x)) = -2
Simplify the equation:
7cos(2x) - 2 + 2cos^2(2x) = -2
7cos(2x) + 2cos^2(2x) = 0
Step 8: Factor out cos(2x)
Factor out cos(2x) from the equation:
cos(2x)(7 + 2cos(2x)) = 0
Step 9: Set each factor equal to zero
Set cos(2x) = 0:
cos(2x) = 0
Set 7 + 2cos(2x) = 0 and solve for cos(2x):
7 + 2cos(2x) = 0
2cos(2x) = -7
cos(2x) = -7/2 (This does not produce any solutions in the given range.)
Step 10: Solve for x
To solve for x, you need to solve for 2x first. Let's solve cos(2x) = 0:
cos(2x) = 0
Use the inverse cosine function (arccos) to solve for 2x:
2x = arccos(0)
2x = π/2 (This is for the first solution.)
Since the range is [0, 2π], we need to find the second solution in the range. For the second solution, we can use the symmetry of the cosine function.
Since cos(2x) = 0, for the second solution, 2x = π - (π/2) = π/2.
To convert 2x back to x, divide by 2:
x = π/4
Therefore, the solutions to the equation 2(sinx)^2 - 5cosx - 4 = 0 between [0, 2π] are x = π/4 and x = π/2.