Solving the Trigonometric Equation for Angle x
We are given the equation:
$3\cos^2 x - 2\sin^2 x = -0.75$
And the constraint $0^\circ \le x \le 90^\circ$.
Simplifying the Equation
To solve for $x$, we can express the equation in terms of a single trigonometric function. Using the identity $\sin^2 x + \cos^2 x = 1$, we substitute $\sin^2 x = 1 - \cos^2 x$ into the equation:
$3\cos^2 x - 2(1 - \cos^2 x) = -0.75$
Calculating cos2 x
Now, simplify and solve for $\cos^2 x$:
$3\cos^2 x - 2 + 2\cos^2 x = -0.75$
Combine the $\cos^2 x$ terms:
$5\cos^2 x - 2 = -0.75$
Add 2 to both sides:
$5\cos^2 x = 2 - 0.75$
$5\cos^2 x = 1.25$
Divide by 5:
$\cos^2 x = \frac{1.25}{5}$
$\cos^2 x = 0.25$
Finding the Angle x
Take the square root of both sides:
$\cos x = \sqrt{0.25}$
Since $x$ is in the range $0^\circ \le x \le 90^\circ$, $\cos x$ must be positive.
$\cos x = 0.5$
We know that the angle whose cosine is 0.5 is $60^\circ$.
$x = 60^\circ$
This value falls within the specified range $0^\circ \le x \le 90^\circ$. Therefore, the angle $x$ is $60^\circ$.
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