Ok, i did this maths trigo question, that says sinx-cos=2/3. I squared it on both sides to get sin2x=5/9 and did the sum. In the end i got two solutions, 16.9 and 73.1. However, when i substituted the answers back into the equation, only one of the 2 answers was correct! The other one didn't work out! Why is this so? I'm pretty sure both answers are correct, so what's going on? Please help!
You have done the right step, check your answers after you have squared each side of an equation.
Take a simple case of
x=-4
square both sides,
x²=(-4)²=16
x=√16=±4
check:
x=-4 OK
x=+4 not a solution.
In your case, the solution of 73.126° is correct. The other one is false.
When you squared both sides of the equation sin(x) - cos(x) = 2/3, you introduced extraneous solutions. This means that some of the solutions obtained by squaring both sides may not actually satisfy the original equation.
To understand why this happens, it is essential to note that squaring both sides of an equation can introduce additional solutions that were not present in the original equation. This is because squaring can eliminate the information about the sign of the original equation.
In your case, when you squared both sides, you obtained sin^2(x) = 5/9. However, this equation has more than just the original two solutions. The solutions you found, 16.9 and 73.1, are correct solutions to sin(x) - cos(x) = 2/3. However, when you squared both sides, extraneous solutions crept in.
To verify this, you can substitute each solution back into the original equation sin(x) - cos(x) = 2/3. You will find that one of the solutions satisfies the original equation, while the other one does not. This inconsistency is due to the introduction of extraneous solutions through squaring both sides.
To avoid this issue, it is important to carefully examine the solutions obtained by squaring both sides and verify them by substituting them back into the original equation.
To understand why only one of your solutions is correct, let's go through the steps of solving the trigonometric equation sin(2x) = 5/9.
Step 1: Start with the initial equation sin(x) - cos(x) = 2/3.
Step 2: Square both sides of the equation to eliminate the square root: (sin(x) - cos(x))^2 = (2/3)^2.
Step 3: Expand the squared expression: sin^2(x) - 2sin(x)cos(x) + cos^2(x) = 4/9.
Step 4: Apply the Pythagorean identity, sin^2(x) + cos^2(x) = 1, and rearrange the equation: 1 - 2sin(x)cos(x) = 4/9.
Step 5: Move the constants to the other side: -2sin(x)cos(x) = 4/9 - 1 = -5/9.
Step 6: Divide both sides by -2: sin(x)cos(x) = 5/18.
Step 7: Double-angle identity for sine: sin(2x) = 2sin(x)cos(x).
Step 8: Substitute sin(x)cos(x) from Step 6: sin(2x) = 2 * (5/18) = 10/18 = 5/9.
Now, to find the solutions for sin(2x) = 5/9, let's use inverse trigonometric functions.
Step 9: Take the inverse sine of both sides: 2x = arcsin(5/9).
Step 10: Divide both sides by 2: x = (1/2) * arcsin(5/9).
At this point, let's calculate the values for x using a calculator:
x₁ = (1/2) * arcsin(5/9) ≈ 16.9°
x₂ = (1/2) * (180° - arcsin(5/9)) ≈ 73.1°
Now, let's substitute these values back into the original equation sin(x) - cos(x) = 2/3 to check if both solutions are valid:
For x₁ ≈ 16.9°: sin(16.9°) - cos(16.9°) ≈ 0.666 - 0.526 ≈ 0.14
This value does not match 2/3, so x₁ is not a valid solution for the original equation.
For x₂ ≈ 73.1°: sin(73.1°) - cos(73.1°) ≈ 0.951 - (-0.309) ≈ 1.26
This value also does not match 2/3, so x₂ is not a valid solution.
Therefore, neither of the solutions you obtained, 16.9 and 73.1, satisfy the original equation sin(x) - cos(x) = 2/3. The reason only one solution works is that during the process of squaring both sides of the equation, extraneous solutions may be introduced, meaning solutions that do not satisfy the original equation. In this case, squaring both sides led to the introduction of extraneous solutions.