Five mathematical trigonometric quadratic equations solved with answers
1. Solve for x: cos^2(x) - 3sin(x)cos(x) + 2sin^2(x) = 0
We can rewrite the left side of the equation using the identity cos^2(x) + sin^2(x) = 1 as follows:
cos^2(x) - 3sin(x)cos(x) + 2sin^2(x)
= cos^2(x) + sin^2(x) - 2sin(x)cos(x) - sin^2(x) + sin^2(x) + sin^2(x)
= 1 - 2sin(x)cos(x) - sin^2(x) + sin^2(x) + sin^2(x)
= 1 - sin^2(x) - 2sin(x)cos(x) + 2sin^2(x)
= (1 - sin(x))(1 - 2sin(x))
Therefore, we have two solutions:
1 - sin(x) = 0
sin(x) = 1
or
1 - 2sin(x) = 0
sin(x) = 1/2
Solving for x, we have:
x = π/2 + 2πn, where n is an integer (for sin(x) = 1)
or
x = π/6 + 2πn or 5π/6 + 2πn, where n is an integer (for sin(x) = 1/2)
2. Solve for x: 2cos^2(x) + 3sin(x)cos(x) - sin^2(x) = 0
We can factor the left side of the equation as follows:
2cos^2(x) + 3sin(x)cos(x) - sin^2(x)
= 2cos^2(x) - 2sin(x)cos(x) + 5sin(x)cos(x) - sin^2(x)
= 2cos(x)(cos(x) - sin(x)) + sin(x)(5cos(x) - sin(x))
= (2cos(x) + sin(x))(cos(x) - sin(x))
Therefore, we have two solutions:
2cos(x) + sin(x) = 0
sin(x) = -2cos(x)
or
cos(x) - sin(x) = 0
sin(x) = cos(x)
Solving for x, we have:
x = 7π/4 + 2πn or 11π/4 + 2πn, where n is an integer (for sin(x) = -2cos(x))
or
x = π/4 + 2πn or 5π/4 + 2πn, where n is an integer (for sin(x) = cos(x))
3. Solve for x: 2sin^2(x) - 5cos(x)sin(x) + 2cos^2(x) = 1
We can rewrite the left side of the equation using the identity sin^2(x) + cos^2(x) = 1 as follows:
2sin^2(x) - 5cos(x)sin(x) + 2cos^2(x) - 1
= 2(sin^2(x) - 2cos(x)sin(x) + cos^2(x)) - 1
= 2(sin(x) - cos(x))^2 - 1
Therefore, we have one solution:
2(sin(x) - cos(x))^2 = 2
(sin(x) - cos(x))^2 = 1
sin(x) - cos(x) = ±1
Solving for x, we have:
x = π/4 + 2πn or 5π/4 + 2πn, where n is an integer (for sin(x) - cos(x) = 1)
or
x = 3π/4 + 2πn or 7π/4 + 2πn, where n is an integer (for sin(x) - cos(x) = -1)
4. Solve for x: cos(x/2) - sin(x/2) = 1/3
We can square both sides of the equation to get rid of the square roots:
cos(x/2)^2 - 2sin(x/2)cos(x/2) + sin(x/2)^2 = 1/9
1 - sin(x) + 1 - cos(x) = 1/9
sin(x) + cos(x) = 16/9
We can use the identity sin(x) + cos(x) = √2cos(x - π/4) to get:
√2cos(x - π/4) = 16/9
Solving for x, we have:
x - π/4 = ±cos^-1(8/9√2)
x = π/4 ± cos^-1(8/9√2)
5. Solve for x: 3cos^2(x) - 4cos(x) + 1 = 0
We can factor the left side of the equation as follows:
3cos^2(x) - 4cos(x) + 1
= (3cos(x) - 1)(cos(x) - 1)
Therefore, we have two solutions:
3cos(x) - 1 = 0
cos(x) = 1/3
or
cos(x) - 1 = 0
cos(x) = 1
Solving for x, we have:
x = cos^-1(1/3) ± 2πn, where n is an integer (for cos(x) = 1/3)
or
x = 2πn, where n is an integer (for cos(x) = 1)