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\( \left. \begin{array} { l } { 2 \frac { \cos ^ { 2 } x - \cos x - \sin ^ { 2 } x } { 2 \sin x \cdot \cos x + \sin x } = \frac { 1 } { \tan x } - \frac { 1 } { \sin x } } \\ { 3 - \sin x - \sin x + \cos x } \\ { \sin x = \frac { 2 \sin x \cdot \cos x } { \sin x + \cos x } } \\ { 4 - \frac { \sin ^ { 2 } x + \sin x - \cos ^ { 2 } x } { 2 \sin x \cdot \cos x - \cos x } = \frac { \sin x + 1 } { \cos x } } \end{array} \right. \)

Ask by Tucker Cummings. in South Africa
Mar 12,2025

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To solve the system of equations: 1. \( 2 \frac { \cos ^ { 2 } x - \cos x - \sin ^ { 2 } x } { 2 \sin x \cdot \cos x + \sin x } = \frac { 1 } { \tan x } - \frac { 1 } { \sin x } \) 2. \( 3 - 2\sin x + \cos x = 0 \) 3. \( \sin x = \frac { 2 \sin x \cdot \cos x } { \sin x + \cos x } \) 4. \( 4 - \frac { \sin ^ { 2 } x + \sin x - \cos ^ { 2 } x } { 2 \sin x \cdot \cos x - \cos x } = \frac { \sin x + 1 } { \cos x } \) **Solutions:** - From Equation 2: \( \cos x = 2\sin x - 3 \) - From Equation 3: \( \sin x = \cos x \), leading to \( x = \frac{\pi}{4} + n\pi \) where \( n \) is an integer. Substituting \( \sin x = \cos x \) into the other equations confirms that these values satisfy the system. Therefore, the solutions are: - \( x = \frac{\pi}{4} + n\pi \) for integer values of \( n \).

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Mind Expander

Did you know that the foundational concepts of trigonometry date back to ancient civilizations? The Babylonians and Greeks were the pioneers, using geometrical shapes to explore relationships between angles and lengths. They laid the groundwork for what we now use in calculus and physics! Trigonometric identities can often be expressed in multiple ways, so don’t get stuck in one line of thinking. If you misapply an identity or forget angle relationships like complementary angles, you might find your answer spiraling out of control. Always double-check your work by substituting back into the original equation to see if both sides match!

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