2sin 2x + cos 3x - cos x= 0 sin 4x+ 2cos x = 1 cost - cos 3x = 3 sin x

To solve the given trigonometric equations, we'll tackle them one by one. ### Equation 1: \( 2\sin 2x + \cos 3x - \cos x = 0 \) 1. Rewrite \(\sin 2x\) and \(\cos 3x\) using double angle and triple angle formulas: - \(\sin 2x = 2\sin x \cos x\) - \(\cos 3x = 4\cos^3 x - 3\cos x\) Thus, the equation becomes: \[ 2(2\sin x \cos x) + (4\cos^3 x - 3\cos x) - \cos x = 0 \] This simplifies to: \[ 4\sin x \cos x + 4\cos^3 x - 4\cos x = 0 \] Factor out common terms: \[ 4\cos x(\sin x + \cos^2 x - 1) = 0 \] 2. Set each factor to zero: - \(4\cos x = 0 \Rightarrow \cos x = 0\) Solutions: \(x = \frac{\pi}{2} + n\pi\) for \(n \in \mathbb{Z}\). - \(\sin x + \cos^2 x - 1 = 0 \) Recall \(\cos^2 x = 1 - \sin^2 x\): \[ \sin x + (1 - \sin^2 x) - 1 = 0 \] \[ \sin x - \sin^2 x = 0 \] Factoring gives: \[ \sin x(1 - \sin x) = 0 \] So, \(\sin x = 0\) or \(\sin x = 1\): - \(\sin x = 0\) gives \(x = n\pi\) - \(\sin x = 1\) gives \(x = \frac{\pi}{2} + 2n\pi\) Thus, the solutions for the first equation are: \[ x = n\pi \quad \text{and} \quad x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \] ### Equation 2: \( \sin 4x + 2\cos x = 1 \) 1. Use the identity \( \sin 4x = 2\sin 2x \cos 2x \) to write: \[ 2\sin 2x \cos 2x + 2\cos x = 1 \] 2. Divide the entire equation by 2: \[ \sin 2x \cos 2x + \cos x = \frac{1}{2} \] We can try specific values for \( x \) or manipulate to isolate \( \sin 2x \cos 2x \). 3. Another alternative is to solve in terms of \(\sin\) and \(\cos\): - Rewrite \( \sin 4x \) in terms of \( 2\sin 2x \): - \(\sin 4x = 2\sin 2x \cos 2x\) - Setting this back in the formula might create complex relationships. 4. Solving for specific values might yield more straightforward results. Generally, plotting or numerical analysis can reveal periodic solutions. ### Equation 3: \( \cos t - \cos 3x = 3 \sin x \) 1. Rewrite \(\cos 3x\) as \(4\cos^3 x - 3\cos x\): \[ \cos t - (4\cos^3 x - 3\cos x) = 3\sin x \] This rearranges to: \[ \cos t = 4\cos^3 x - 3\cos x + 3\sin x \] This equation, in Cartesian coordinates or numerical boundaries, can find solutions for both \( \cos t \) and other identities. To conclude, each equation necessitates varied solving approaches either analytically or graphically for broader periodic solutions. Would you like to focus on further exploration of any particular equation?