\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
By the identity, we have \[ \cos x + \cos 4x = 2 \cos \frac {5}{2} x \cos \frac {3}{2}x, \] and \[ \cos 2x + \cos 3x = 2 \cos \frac {5}{2}x \cos \frac {1}{2}x. \]
Hence, we have \[ \cos x + 3 \cos 2x + 3 \cos 3x = 2 \cos \frac {5}{2} x \left (\cos \frac {3}{2} x + 3 \cos \frac {1}{2}x\right ) = 0. \]
Hence, either \[ \cos \frac {5}{2} x = 0, \] or \[ \cos \frac {3}{2} x + 3 \cos \frac {1}{2}x = 0. \]
In the first case, we have \(\frac {5}{2} x = \frac {1}{2}\pi + k\pi \) for \(k \in \ZZ \), and hence \[ x = \frac {1 + 2k}{5} \cdot \pi . \]
Since \(0 \leq x \leq 2\pi \), we have \[ 0 \leq \frac {1 + 2k}{5} \leq 2, \] and hence \[ 0 \leq 1 + 2k \leq 10, \] giving \(k = 0, 1, 2, 3, 4\). Hence, the solutions are \[ x = \frac {1}{5} \pi , x = \frac {3}{5} \pi , x = \pi , x = \frac {7}{5} \pi , x = \frac {9}{5}\pi . \]
In the second case, notice that \begin {align*} \cos 3t & = \cos (2t + t) \\ & = \cos 2t \cos t - \sin 2t \sin t \\ & = (\cos ^2 t - \sin ^2 t) \cos t - 2 \sin ^2 t \cos t \\ & = \cos ^3 t - 3 \sin ^2 \cos t. \end {align*}
Hence, \[ \cos \frac {3}{2}x + 3\cos \frac {1}{2} x = 0 \iff \cos ^3 \frac {1}{2}x - 3 \sin ^2 \frac {1}{2}x \cos \frac {1}{2}x + 3 \cos \frac {1}{2}x = 0, \] and using the identity \(\sin ^2 t + \cos ^2 t = 1\), this simplifies to \[ \cos ^3 \frac {1}{2}x + 3 \cos ^3 \frac {1}{2}x = 0, \] which is \[ \cos \frac {1}{2}x = 0. \]
This gives \[ \frac {1}{2}x = \frac {\pi }{2} + k\pi \] for \(k \in \ZZ \), and hence \[ x = (1 + 2k) \pi . \]
Since \(0 \leq x \leq 2\pi \), the only \(k\) valid is \(k = 0\), and this solves to \(x = \pi \).
Hence, all the solutions to this equation is \[ x \in \left \{\frac {1}{5} \pi , \frac {3}{5} \pi , \pi , \frac {7}{5} \pi , \frac {9}{5}\pi \right \}. \]
Using the given identity, we have \[ \cos (x + y) + \cos (x - y) = 2 \cos x \cos y. \]
Hence, the original equation simplifies to \[ 2 \cos x \cos y - \cos 2x = 1. \]
Using the identity \(\cos 2x = 2 \cos ^2 x - 1\), and this gives \[ 2 \cos x \cos y - (2 \cos ^2 x - 1) = 1, \] and hence \[ 2 \cos x \cos y - 2 \cos ^2 x = 0, \] which means \[ \cos x (\cos y - \cos x) = 0, \] and hence \(\cos x = 0\) or \(\cos y - \cos x = 0\).
The first one gives us \(x = \frac {\pi }{2}\) in the range \(x \in [0, \pi ]\).
Since \(\cos \) is one-to-one when restricted to \([0, \pi ]\), the second one is equivalent to \(\cos y = \cos x\) which is equivalent to \(x = y\).
The specific value is \(x = \frac {\pi }{2}\).
Using the identity given, we have \[ \cos x + \cos y = 2 \cos \frac {x + y}{2} \cos \frac {x - y}{2}, \] and \[ \cos (x + y) = 2 \cos ^2 \frac {x + y}{2} - 1. \]
Let \(u = \frac {x + y}{2}\) and \(v = \frac {x - y}{2}\). We have \(0 \leq u \leq \pi \) and \(-\frac {\pi }{2} \leq v \leq \frac {\pi }{2}\), and the original equation simplifies to \[ 2 \cos u \cos v - 2 \cos ^2 u + 1 = \frac {3}{2}, \] and hence \[ 4 \cos u \cos v - 4 \cos ^2 u + 2 = 3, \] and \[ 4 \cos ^2 u - 4 \cos u \cos v + 1 = 0. \]
Since \(1 = \cos ^2 v + \sin ^2 v\), we have \[ 4 \cos ^2 u - 4 \cos u \cos v + \cos ^2 v = - \sin ^2 v, \] and hence \[ (2 \cos u - \cos v)^2 = - \sin ^2 v. \]
The left-hand side is non-negative, and the right-hand side is non-positive. Hence, the only way for the equal sign to take place is when both sides are zero, which is \[ 2 \cos u = \cos v, \sin v = 0. \]
Within this range of \(v\), the only case where \(\sin v = 0\) is when \(v = 0\), and hence \(2 \cos u = 1\), \(\cos u = \frac {1}{2}\), leading to \(u = \frac {\pi }{3}\).
Hence, \(x = u + v = \frac {\pi }{3}\), and \(y = u - v = \frac {\pi }{3}\), and the only solution is \[ (x, y) = \left (\frac {\pi }{3}, \frac {\pi }{3}\right ). \]