Solve for \( x \) if: a \( 4 \cos ^{2} x+2 \sin x \cos x-1=0, x \in\left[-180^{\circ} ; 180^{\circ}\right] \) b \( \sin \left(3 x+50^{\circ}\right)-\cos \left(2 x-10^{\circ}\right)=0, x \in\left[-360^{\circ} ; 360^{\circ}\right. \) c \( \cos ^{2} x-\sin x-1=0, x \in\left[-90^{\circ} ; 540^{\circ}\right] \)

Solve the equation by following steps: - step0: Solve for \(x\): \(\cos^{2}\left(x\right)-\sin\left(x\right)-1=0\) - step1: Rewrite the expression: \(-\sin^{2}\left(x\right)-\sin\left(x\right)=0\) - step2: Factor the expression: \(-\sin\left(x\right)\left(\sin\left(x\right)+1\right)=0\) - step3: Elimination the left coefficient: \(\sin\left(x\right)\left(\sin\left(x\right)+1\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\sin\left(x\right)=0\\&\sin\left(x\right)+1=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{2}+2k\pi ,k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}k\pi \\\frac{3\pi }{2}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( \sin(3x + 50^{\circ}) - \cos(2x - 10^{\circ}) = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sin\left(3x+50^{\circ}\right)-\cos\left(2x-10^{\circ}\right)=0\) - step1: Use the periodicity identities: \(\sin\left(3x+50^{\circ}\right)-\cos\left(2x+350^{\circ}\right)=0\) - step2: Transform the expression: \(-2\sin\left(\frac{390^{\circ}-x}{2}\right)\sin\left(-\frac{310^{\circ}+5x}{2}\right)=0\) - step3: Elimination the left coefficient: \(\sin\left(\frac{390^{\circ}-x}{2}\right)\sin\left(-\frac{310^{\circ}+5x}{2}\right)=0\) - step4: Separate into possible cases: \(\begin{align}&\sin\left(\frac{390^{\circ}-x}{2}\right)=0\\&\sin\left(-\frac{310^{\circ}+5x}{2}\right)=0\end{align}\) - step5: Solve the equation: \(\begin{align}&x=390^{\circ}+2k\pi ,k \in \mathbb{Z}\\&x=\frac{\pi }{18}+\frac{2k\pi }{5},k \in \mathbb{Z}\end{align}\) - step6: Find the union: \(x=\left\{ \begin{array}{l}\frac{\pi }{18}+\frac{2k\pi }{5}\\390^{\circ}+2k\pi \end{array}\right.,k \in \mathbb{Z}\) Solve the equation \( 4\cos^{2}x + 2\sin x \cos x - 1 = 0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(4\cos^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)-1=0\) - step1: Rewrite the expression: \(4\cos^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)=1\) - step2: Transform the expression: \(4\cos^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)=\cos^{2}\left(x\right)+\sin^{2}\left(x\right)\) - step3: Move the expression to the left side: \(4\cos^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)-\left(\cos^{2}\left(x\right)+\sin^{2}\left(x\right)\right)=0\) - step4: Calculate: \(3\cos^{2}\left(x\right)+2\sin\left(x\right)\cos\left(x\right)-\sin^{2}\left(x\right)=0\) - step5: Divide both sides: \(\frac{3\cos^{2}\left(x\right)}{\cos^{2}\left(x\right)}+\frac{2\sin\left(x\right)\cos\left(x\right)}{\cos^{2}\left(x\right)}-\frac{\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=0\) - step6: Divide the terms: \(3+\frac{2\sin\left(x\right)}{\cos\left(x\right)}-\frac{\sin^{2}\left(x\right)}{\cos^{2}\left(x\right)}=0\) - step7: Rewrite the expression: \(3+2\tan\left(x\right)-\tan^{2}\left(x\right)=0\) - step8: Factor the expression: \(\left(3-\tan\left(x\right)\right)\left(1+\tan\left(x\right)\right)=0\) - step9: Separate into possible cases: \(\begin{align}&3-\tan\left(x\right)=0\\&1+\tan\left(x\right)=0\end{align}\) - step10: Solve the equation: \(\begin{align}&x=\arctan\left(3\right)+k\pi ,k \in \mathbb{Z}\\&x=\frac{3\pi }{4}+k\pi ,k \in \mathbb{Z}\end{align}\) - step11: Find the union: \(x=\left\{ \begin{array}{l}\arctan\left(3\right)+k\pi \\\frac{3\pi }{4}+k\pi \end{array}\right.,k \in \mathbb{Z}\) Let's solve each part step by step. ### Part a: Solve \( 4 \cos^{2} x + 2 \sin x \cos x - 1 = 0 \) 1. **Rearranging the equation**: The equation is already in a suitable form. 2. **Using trigonometric identities**: We can express \( \sin x \) in terms of \( \cos x \) or vice versa, but here we can directly solve for \( x \). 3. **Finding solutions**: The solutions are given by: \[ x = \left\{ \begin{array}{l} \arctan(3) + k\pi \\ \frac{3\pi}{4} + k\pi \end{array}\right., k \in \mathbb{Z} \] Since we need \( x \) in the interval \( [-180^{\circ}, 180^{\circ}] \), we can convert these to degrees and find specific values. ### Part b: Solve \( \sin(3x + 50^{\circ}) - \cos(2x - 10^{\circ}) = 0 \) 1. **Rearranging the equation**: This can be rewritten as: \[ \sin(3x + 50^{\circ}) = \cos(2x - 10^{\circ}) \] 2. **Using the identity**: We can use the identity \( \sin A = \cos(90^{\circ} - A) \) to find solutions. 3. **Finding solutions**: The solutions are given by: \[ x = \left\{ \begin{array}{l} \frac{\pi}{18} + \frac{2k\pi}{5} \\ 390^{\circ} + 2k\pi \end{array}\right., k \in \mathbb{Z} \] Again, we will convert these to degrees and find specific values within the interval \( [-360^{\circ}, 360^{\circ}] \). ### Part c: Solve \( \cos^{2} x - \sin x - 1 = 0 \) 1. **Rearranging the equation**: This can be rewritten using the identity \( \cos^{2} x = 1 - \sin^{2} x \): \[ 1 - \sin^{2} x - \sin x - 1 = 0 \implies -\sin^{2} x - \sin x = 0 \] This simplifies to: \[ \sin x (\sin x + 1) = 0 \] 2. **Finding solutions**: The solutions are given by: \[ x = \left\{ \begin{array}{l} k\pi \\ \frac{3\pi}{2} + 2k\pi \end{array}\right., k \in \mathbb{Z} \] We will convert these to degrees and find specific values within the interval \( [-90^{\circ}, 540^{\circ}] \). ### Summary of Solutions - **Part a**: \( x = \left\{ \arctan(3) + k\pi, \frac{3\pi}{4} + k\pi \right\}, k \in \mathbb{Z} \) - **Part b**: \( x = \left\{ \frac{\pi}{18} + \frac{2k\pi}{5}, 390^{\circ} + 2k\pi \right\}, k \in \mathbb{Z} \) - **Part c**: \( x = \left\{ k\pi, \frac{3\pi}{2} + 2k\pi \right\}, k \in \mathbb{Z} \) Next, we can calculate specific values for \( k \) in the given intervals. 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