Solve 4sin^2x + 4(squareroot of 2)cosx -6 for all real values of x.
Did you mean
4sin^2x + 4√2cosx -6 = 0 ?
if so then
4(1-cos^2x) + 4√2cosx - 6 = 0
4 - cos^2x + 4√2cosx - 6 = 0
2cos^2x + 2√2cosx + 1 = 0
let cosx = y
then y = (-2√2 ± √(8 - 4(2)(1))/4
y = -√2/2
then cosx = -√2/2
we know that cos 45 = √2/2
and x must be in the II or III quadrant, so
x = 180-45 = 135 degrees or
x = 180+45 = 225 degrees
in radians that would be 3pi/4 or 5pi/4 radians
general solution
x = 135 + 360k, 225 + 360k, k an integer
or
x = 3pi/4 + 2kpi, 5pi/4 + 2kpi
Thank you for helping me with this problem.
Joanie
To solve the equation 4sin^2x + 4√2cosx - 6 = 0 for all real values of x, we can try to simplify it using trigonometric identities.
First, notice that the coefficient of sin^2x is 4, which is twice the coefficient of cos^2x. This suggests that we can use the double angle formula for sine to express sin^2x in terms of cosx.
The double angle formula for sine states that sin^2x = (1/2)(1 - cos2x). Substituting this into the equation, we have:
4(1/2)(1 - cos2x) + 4√2cosx - 6 = 0
Simplifying further:
2(1 - cos2x) + 4√2cosx - 6 = 0
2 - 2cos2x + 4√2cosx - 6 = 0
-2cos2x + 4√2cosx - 4 = 0
Now, let's try to factor out a common term. Notice that the equation contains only terms involving cosx. Factoring out 2cosx, we have:
2cosx(-cosx + 2√2) - 4 = 0
Next, we can divide the entire equation by 2 to simplify a bit further:
cosx(-cosx + 2√2) - 2 = 0
Now, let's focus on the expression in the parentheses. We notice that we have a quadratic-like expression: -cosx + 2√2. To solve it, we can set it equal to zero:
-cosx + 2√2 = 0
Adding cosx to both sides:
2√2 = cosx
So, cosx = 2√2
To find the values of x that satisfy this equation, we can apply the inverse cosine function (also known as arccos or cos^(-1)) to both sides:
x = arccos(2√2)
Now, the inverse cosine function returns a value between 0 and π, or between 0 and 180 degrees. Therefore, the equation x = arccos(2√2) gives us the values of x within this range that satisfy the equation.
To summarize, the solution to the equation 4sin^2x + 4√2cosx - 6 = 0 for all real values of x is given by x = arccos(2√2), where x lies between 0 and π (or between 0 and 180 degrees).