Solve 4sin^2 x + 4sqrt(2) cos x-6 = 0 for all real values of x.
My answer: (pi/4)+2(pi)k, (7pi/4) + 2(pi)k
Solve 2cos^2 x-5cosx+2=0 for principal values of x.
My answer: 60 and 300 degrees
Very good, I agree with both answers.
Is there any reason why you gave the first answers in radians and the second in degrees?
To solve the equation 4sin^2(x) + 4sqrt(2)cos(x) - 6 = 0 for all real values of x, let's follow these steps:
Step 1: Rearrange the equation to group the sine and cosine terms separately:
4sin^2(x) + 4sqrt(2)cos(x) - 6 = 0
Rewrite sqrt(2)cos(x) as sqrt(2) * cos(x) to separate the terms:
4sin^2(x) + 4sqrt(2) * cos(x) - 6 = 0
Step 2: Factor out a common 4 from the first two terms:
4(sin^2(x) + sqrt(2) * cos(x)) - 6 = 0
Step 3: Complete the square for the terms inside the parentheses:
(sin(x) + sqrt(2)/2)^2 - (sqrt(2)/2)^2 + 4sqrt(2) * cos(x) - 6 = 0
(sin(x) + sqrt(2)/2)^2 - 2 + 4sqrt(2) * cos(x) - 6 = 0
(sin(x) + sqrt(2)/2)^2 + 4sqrt(2) * cos(x) - 8 = 0
Step 4: Notice that we can rewrite the cosine term as 2 * sqrt(2) * cos(x):
(sin(x) + sqrt(2)/2)^2 + 2 * sqrt(2) * 2 * cos(x) - 8 = 0
(sin(x) + sqrt(2)/2)^2 + 4 * sqrt(2) * cos(x) - 8 = 0
Step 5: Substitute a new variable to simplify the equation further. Let's say u = sin(x) + sqrt(2)/2:
u^2 + 4sqrt(2) * cos(x) - 8 = 0
Step 6: We can now solve the quadratic equation for u. By factoring or using the quadratic formula, we find:
(u - 2sqrt(2))(u + 2sqrt(2)) = 0
Step 7: Solve for u in each factor separately:
u - 2sqrt(2) = 0 or u + 2sqrt(2) = 0
Step 8: Solve for sin(x) in each equation by substituting back the value of u:
sin(x) + sqrt(2)/2 = 2sqrt(2) or sin(x) + sqrt(2)/2 = -2sqrt(2)
Step 9: Solve for sin(x) separately:
sin(x) = 2sqrt(2) - sqrt(2)/2 or sin(x) = -2sqrt(2) - sqrt(2)/2
Step 10: Simplify the values of sin(x):
sin(x) = (4sqrt(2) - sqrt(2))/2 or sin(x) = (-4sqrt(2) - sqrt(2))/2
sin(x) = 3sqrt(2)/2 or sin(x) = -5sqrt(2)/2
Step 11: Find the values of x by taking the inverse sine of each value:
x = arcsin(3sqrt(2)/2) or x = arcsin(-5sqrt(2)/2)
Step 12: Since we want to find all real values of x, we can use the periodicity of the sine function to find the general solutions. The general solution for x can be represented as:
x = (pi/4) + 2(pi)k or x = (7pi/4) + 2(pi)k, where k is an integer.
Therefore, the solutions for the equation 4sin^2(x) + 4sqrt(2)cos(x) - 6 = 0 for all real values of x are:
x = (pi/4) + 2(pi)k or x = (7pi/4) + 2(pi)k, where k is an integer.
Now let's solve the equation 2cos^2(x) - 5cos(x) + 2 = 0 for the principal values of x:
Step 1: Let's factor the quadratic equation:
(2cos(x) - 1)(cos(x) - 2) = 0
Step 2: Set each factor equal to zero and solve for cos(x):
2cos(x) - 1 = 0 or cos(x) - 2 = 0
Step 3: Solve for cos(x) in each equation:
cos(x) = 1/2 or cos(x) = 2
Step 4: Find the principal values of x by taking the inverse cosine of each value:
x = arccos(1/2) or x = arccos(2)
Step 5: Simplify the values of x:
x = 60 degrees or x = 120 degrees
Therefore, the principal values of x for the equation 2cos^2(x) - 5cos(x) + 2 = 0 are:
x = 60 degrees or x = 120 degrees.