Solve for x:
sin2x + cos^2x = 1
0° < or equal to x < 360°
That is an identity, it is true for all values of x.
Try x=-439
o____O
how did you get to that value? can you show me how to solve it algabraeically?
Draw a right triangle with hypotenuse c and sides a and b. Opposite angles are then A, B, C
now sin A = a/c so a = c sin A
cos a = b/c so b = c cos A
right triangle geometry says
a^2 + b^2 =c ^2
so
c^2 sin^2 A + c^2 cos^2 A = c^2
divide both sides by c^2
To solve the equation sin(2x) + cos^2(x) = 1, we can use trigonometric identities to simplify and solve for x. Here's how:
Step 1: Rewrite the equation using the double-angle identity for sine.
sin(2x) = 2sin(x)cos(x)
The equation becomes: 2sin(x)cos(x) + cos^2(x) = 1
Step 2: Combine like terms.
cos^2(x) + 2sin(x)cos(x) - 1 = 0
Step 3: Let's look at this equation as a quadratic equation in terms of cos(x).
Let cos(x) be represented as t.
t^2 + 2sin(x)t - 1 = 0
Step 4: Solve the quadratic equation.
Using the quadratic formula, with a = 1, b = 2sin(x), and c = -1, we get:
t = (-2sin(x) ± √(4sin^2(x) + 4))/2
t = -sin(x) ± √(sin^2(x) + 1)
Step 5: Solve for sin(x) using the identity cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - sin^2(x)
1 - t^2 = 1 - sin^2(x)
sin^2(x) = t^2
Taking the square root of both sides yields:
sin(x) = ± |t|
Step 6: Solve for x.
Since we know that 0° ≤ x < 360°, we'll need to find the possible values of x that satisfy the equation.
From Step 5, we know that sin(x) = ± |t|.
Since -1 ≤ t ≤ 1, sin(x) can only be between -1 and 1.
So, we have two possibilities:
1) If sin(x) = t:
sin(x) = t
sin(x) = ± t
2) If sin(x) = -t:
sin(x) = -t
x = arcsin(-t)
Now, let's substitute t back in terms of cos(x).
Recall t = -sin(x) ± √(sin^2(x) + 1).
For the first case:
sin(x) = -sin(x) ± √(sin^2(x) + 1)
2sin(x) ± √(sin^2(x) + 1) = 0
(since sin(x) cannot be 0)
Solving for sin(x), we get:
sin(x) = ± √(-1)
which has no real solutions.
For the second case:
sin(x) = -t
sin(x) = sin(x) ± √(sin^2(x) + 1)
2sin(x) ± √(sin^2(x) + 1) = 0
Simplifying, we get:
√(sin^2(x) + 1) = -2sin(x)
Since the square root on the left-hand side is always positive, for this equation to hold, the expression on the right-hand side must also be non-negative. Therefore, we can ignore the negative sign:
2sin(x) + √(sin^2(x) + 1) = 0
Squaring both sides:
(2sin(x))^2 = (√(sin^2(x) + 1))^2
4sin^2(x) = sin^2(x) + 1
3sin^2(x) = 1
Finally, solving for sin(x):
sin^2(x) = 1/3
sin(x) = ± √(1/3)
Since we know that sin(x) = ± √(1/3), we can find the possible values of x within the given range using the inverse sine function (arcsin). Using a calculator, we find:
x = arcsin(√(1/3)) ≈ 30°
x = 180° - arcsin(√(1/3)) ≈ 150°
x = 180° + arcsin(√(1/3)) ≈ 210°
x = 360° - arcsin(√(1/3)) ≈ 330°
So, the solution to the equation sin(2x) + cos^2(x) = 1, where 0° ≤ x < 360°, is:
x ≈ 30°, 150°, 210°, 330°