The values of x in 2cos(2x) = 11cos(x) + 1 are x = 1.82 and x = 4.46
How to solve for x?
The equation is given as:
2cos(2x) = 11cos(x) + 1
Expand cos(2x)
2(cos²(x) - sin²(x)) = 11cos(x) + 1
sin²(x) = 1 - cos²(x)
So, we have:
2(cos²(x) - 1 + cos²(x)) = 11cos(x) + 1
Evaluate the like terms
2(2cos²(x) - 1) = 11cos(x) + 1
Expand
4cos²(x) - 2 = 11cos(x) + 1
Rewrite as:
4cos²(x) - 11cos(x) - 2 - 1 = 0
4cos²(x) - 11cos(x) - 3 = 0
Let cos(x) = y.
So, we have
4y² - 11y - 3 = 0
Expand
4y² + y - 12y - 3 = 0
Factorize
y(4y + 1) - 3(4y + 1) = 0
Factor out 4y + 1
(y - 3)(4y + 1) = 0
Solve for y
y = 3 or 4y = -1
This gives
y = 3 or y = -1/4
Recall that y = cos(x)
So, we have:
cos(x) = 3 or cos(x) = -1/4
The cosine of x is always between -1 and 1 (inclusive).
So, we have:
cos(x) = -1/4
Take the arc cos
[tex]x =\cos^{-1}(-1/4)[/tex]
Evaluate the arccos (in radians)
x = 1.82 and x = 4.46
Hence, the values of x in 2cos(2x) = 11cos(x) + 1 are x = 1.82 and x = 4.46
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