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

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**(a)** We start with
$$
7\cos^2x+\sin x\cos x-6=0.
$$
It will be helpful to write everything in terms of double–angle functions. Recall that
$$
\cos^2x=\frac{1+\cos2x}{2}\quad\text{and}\quad \sin x\cos x=\frac{\sin2x}{2}.
$$
Substitute these into the equation:
$$
7\left(\frac{1+\cos2x}{2}\right)+\frac{\sin2x}{2}-6=0.
$$
Multiply both sides by 2 to eliminate the denominator:
$$
7(1+\cos2x)+\sin2x-12=0.
$$
Simplify:
$$
7\cos2x+\sin2x-5=0,
$$
or equivalently,
$$
7\cos2x+\sin2x=5.
$$

We now express the left‐side as a single cosine function. Write
$$
7\cos2x+\sin2x=R\cos(2x-\phi),
$$
where $$R>0$$ and $$\phi$$ is an angle to be determined. Expanding the right side using the cosine subtraction formula we get
$$
R\cos2x\cos\phi+R\sin2x\sin\phi.
$$
Thus, we require
$$
R\cos\phi=7\quad\text{and}\quad R\sin\phi=1.
$$
Then
$$
R=\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt{2},
$$
and
$$
\phi=\arctan\left(\frac{1}{7}\right).
$$
The equation becomes
$$
5\sqrt{2}\cos\bigl(2x-\arctan(1/7)\bigr)=5.
$$
Divide both sides by $$5\sqrt{2}$$:
$$
\cos\Bigl(2x-\arctan(1/7)\Bigr)=\frac{1}{\sqrt{2}}.
$$

We now solve
$$
\cos\theta=\frac{1}{\sqrt{2}},
$$
with
$$
\theta=2x-\arctan(1/7).
$$
The general solution for $$\theta$$ is
$$
\theta=2\pi k\pm \frac{\pi}{4},\quad k\in\mathbb{Z}.
$$
That is,
$$
2x-\arctan(1/7)=2\pi k\pm\frac{\pi}{4}.
$$
Solving for $$x$$:
$$
2x=\arctan(1/7)+2\pi k\pm \frac{\pi}{4},
$$
$$
x=\frac{1}{2}\left(\arctan(1/7)\pm\frac{\pi}{4}\right)+\pi k,\quad k\in\mathbb{Z}.
$$

---

**(b)** We need to solve
$$
\cos^2x-2\cos x=0.
$$
Factor out $$\cos x$$:
$$
\cos x\left(\cos x-2\right)=0.
$$
Thus we have two cases:

1. $$\cos x=0$$, or
2. $$\cos x-2=0$$ (i.e. $$\cos x=2$$).

Since the cosine function always satisfies $$-1\le \cos x\le 1$$, $$\cos x=2$$ has no solution. Therefore, only
$$
\cos x=0
$$
remains.

The general solution for $$\cos x=0$$ is
$$
x=\frac{\pi}{2}+\pi k,\quad k\in\mathbb{Z}.
$$

---

**Final Answers:**

(a) $$\displaystyle x=\frac{1}{2}\left(\arctan\left(\frac{1}{7}\right)\pm\frac{\pi}{4}\right)+\pi k,\quad k\in\mathbb{Z}$$.

(b) $$\displaystyle x=\frac{\pi}{2}+\pi k,\quad k\in\mathbb{Z}$$.
**(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: $$ x = \frac{\pi}{2} + \pi k,\quad k \in \mathbb{Z} $$

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

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