Solve the equation 2cos2x = √3 for 0°≤x≤360°
I did this:
cos2x = √3 /2
2x=30
x=15
x=15, 165, 195, 345
Is this correct?
Solve the equation √3 sin2x + cos2x = 0 for -π≤x≤π
No idea how to approach this one
Thanks a bunch for the help!
2cos2x = √3
cos2x = √3/2
2x = 30°,330° + 360n
x = 14,165 + 180n
x = 15°,195°,165°,345°
You are correct
√3 sin2x + cos2x = 0
√3 sin2x = -cos2x
tan2x = -1/√3
tan < 0 in QII,QIV
and so on as in the other problem.
For the equation 2cos2x = √3, you have made a mistake in the solution.
Here's the correct approach:
Starting from 2cos2x = √3, divide both sides by 2 to get cos2x = √3 / 2.
Now, recall that the cosine function has its values equal to √3 / 2 at two specific angles, which are 30° and 330°.
To find the values of x, we need to consider the double angle formula: cos2x = 2cos^2(x) - 1.
Plugging in cos2x = √3 / 2, we get 2cos^2(x) - 1 = √3 / 2.
Rearranging the equation, we have 2cos^2(x) = (√3 + 1) / 2.
Dividing both sides by 2, we have cos^2(x) = (√3 + 1) / 4.
Taking the square root of both sides, we get cos(x) = ±√[(√3 + 1) / 4].
Now, we can find the values of x by taking the inverse cosine of these values. However, we also need to consider the range of x (0° ≤ x ≤ 360°).
To find the values of x within this range, we can use the symmetry of the cosine function across the y-axis.
The positive square root (√[(√3 + 1) / 4]) will give us the values of x in the first and fourth quadrants, while the negative square root (-√[(√3 + 1) / 4]) will give us the values in the second and third quadrants.
Using a calculator, find the inverse cosine of the positive square root to get the principal value of x, which is approximately 43.3°.
Since the cosine function repeats its values every 360°, we can find the other values of x by adding or subtracting multiples of 360°.
Therefore, the solutions for x are 43.3°, 316.7° (360° - 43.3°), 136.7° (180° + 43.3°), and 223.3° (360° - 136.7°).
So, the correct solution for 0° ≤ x ≤ 360° is x = 43.3°, 136.7°, 223.3°, and 316.7°.
Now, let's move on to the second equation: √3 sin2x + cos2x = 0.
To approach this equation, we need to simplify it further.
Using the double angle formula for the sine function, sin2x = 2sin(x)cos(x), we can rewrite the equation as:
√3(2sin(x)cos(x)) + cos2x = 0.
Simplifying further, we have 2√3sin(x)cos(x) + cos2x = 0.
Now, we can factor out a common term of cos(x):
cos(x)(2√3sin(x) + cos(x)) = 0.
From this equation, we see that either cos(x) = 0 or 2√3sin(x) + cos(x) = 0.
For cos(x) = 0, the values of x can be found by taking the inverse cosine of 0, which is x = 90° and x = 270°.
For 2√3sin(x) + cos(x) = 0, we can rearrange the equation as follows:
2√3sin(x) = - cos(x).
Dividing both sides by 2 and taking the square of both sides, we get:
12sin^2(x) = cos^2(x).
Rearranging the equation, we have:
12sin^2(x) - cos^2(x) = 0.
Using the identity sin^2(x) + cos^2(x) = 1, we can rewrite this equation as:
12sin^2(x) - (1 - sin^2(x)) = 0.
Simplifying further, we have:
12sin^2(x) - 1 + sin^2(x) = 0.
Combining like terms, we get:
13sin^2(x) - 1 = 0.
Adding 1 to both sides, we have:
13sin^2(x) = 1.
Dividing both sides by 13, we get:
sin^2(x) = 1/13.
Taking the square root of both sides, we have:
sin(x) = ±√(1/13).
Now, we can find the values of x by taking the inverse sine of these values. However, we also need to consider the range of x (-π ≤ x ≤ π).
Using a calculator, find the inverse sine of the positive square root (√(1/13)) to get the principal value of x, which is approximately 0.2425 radians or 13.88°.
Since the sine function repeats its values every 2π radians or 360°, we can find the other values of x by adding or subtracting multiples of 2π.
Therefore, the solutions for the equation 2√3sin(x) + cos(x) = 0 are x = 0.2425 radians or 13.88°, 2.8989 radians or 166.12°, and -2.8989 radians or -166.12°.
So, the correct solutions for -π ≤ x ≤ π are x = 0.2425 radians or 13.88°, 2.8989 radians or 166.12°, and -2.8989 radians or -166.12°.
I hope this helps! Let me know if you have any further questions.
To solve the equation 2cos2x = √3 for 0° ≤ x ≤ 360°, you've made a calculation error in your solution.
Begin by dividing both sides of the equation by 2:
cos2x = √3/2
Now, we need to find the values of x within the given range for which the cosine of 2x is equal to √3/2. Recall the unit circle, where the cosine function is positive in the first and fourth quadrants.
In the first quadrant (0° ≤ x ≤ 90°), we have:
2x = 30°
x = 15°
In the fourth quadrant (270° ≤ x ≤ 360°), we have:
2x = 330°
x = 165°
To find the other solutions, we should add multiples of 180° to the original solutions:
x = 15°, 165°, 195°, 345°
Therefore, your solution is correct: x = 15°, 165°, 195°, 345°.
Moving on to the second question:
To solve the equation √3 sin2x + cos2x = 0 for -π ≤ x ≤ π, we need to use trigonometric identities to transform the equation into a solvable form.
Start by using the Pythagorean identity sin²x + cos²x = 1:
√3 sin²x + (1 - sin²x) = 0
Simplify:
√3 sin²x + 1 - sin²x = 0
Combine like terms:
(√3 - 1)sin²x + 1 = 0
Now, we can isolate sin²x:
(√3 - 1)sin²x = -1
sin²x = -1 / (√3 - 1)
From the equation, it's clear that sin²x cannot be negative, so there are no solutions within the given range. Therefore, the equation has no solutions for -π ≤ x ≤ π.
I hope this explanation helps! Let me know if you have any further questions.