The solution of the given trigonometric expression 2.cos(2x) = 7.cos(x), in the range 0° < x < 360°, is x = 104.5° or 255.5°.
In the question, we are given the trigonometric equation 2.cos(2x) = 7.cos(x), and are asked to solve for x, given 0° < x < 360°.
By trigonometric identities, we know that, cos(2θ) = 2.cos²(θ) - 1.
Using this identity, we solve the given expression as:
2.cos(2x) = 7.cos(x),
or, 2{2.cos²(x) - 1} = 7.cos(x),
or, 4.cos²(x) - 2 - 7.cos(x) = 0.
We substitute cos(x) as k, to get a quadratic equation:
4k² - 7k - 2 = 0.
This equation can be solved as:
4k² - 7k - 2 = 0,
or, 4k² - 8k + k - 2 = 0,
or, 4k(k - 2) + 1(k - 2) = 0,
or, (4k + 1)(k - 2) = 0.
By the zero-product rule, we can show that,
Either, 4k + 1 = 0, ⇒ k = -1/4,
or, k - 2 = 0, or, k = 2.
Since k represents cos(x), it cannot have a value of 2 as we know that -1 ≤ cos(θ) ≤ 1.
Thus, we have cos(x) = k = -1/4.
Using this, we can write that
x = cos⁻¹(-1/4) = 104.5° or 255.5° in the range 0° < x < 360°.
Thus, the solution of the given trigonometric expression 2.cos(2x) = 7.cos(x), in the range 0° < x < 360°, is x = 104.5° or 255.5°.
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