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Solve for \( x \) (a) \( 7 \cos ^{2} x+\sin x \cos x-6=0 \) b) \( \cos ^{2} x-2 \cos x=0 \)

Ask by Lang Harrington. in South Africa
Mar 12,2025

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**(a)** The solutions for \( x \) are: \[ x = \frac{1}{2}\left(\arctan\left(\frac{1}{7}\right) \pm \frac{\pi}{4}\right) + \pi k,\quad k \in \mathbb{Z} \] **(b)** The solutions for \( x \) are: \[ x = \frac{\pi}{2} + \pi k,\quad k \in \mathbb{Z} \]

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Bonus Knowledge

To solve the equations, let's tackle them one by one! For \( 7 \cos^2 x + \sin x \cos x - 6 = 0 \): First, recall that \( \sin x = \sqrt{1 - \cos^2 x} \). Let \( y = \cos x \). Then \( \sin x = \sqrt{1 - y^2} \): 1. Substitute and we get: \[ 7y^2 + \sqrt{1 - y^2} \cdot y - 6 = 0 \] This is a bit complex, so let’s backtrack and solve directly using \( \sin x = \sqrt{1 - y^2} \) through numerical methods or graphing to find values of \( x \). For \( \cos^2 x - 2 \cos x = 0 \): 1. Factor out a \( \cos x \): \[ \cos x (\cos x - 2) = 0 \] So, \( \cos x = 0 \) or \( \cos x = 2 \). Since cosine cannot exceed 1, we discard \( \cos x = 2 \). 2. Therefore, \( \cos x = 0 \) leads to: \[ x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \] So, for part (a), numerical or graphical methods will help find the values of \( x \), while for part (b), we simply arrive at \( x = \frac{\pi}{2} + n\pi \).

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