Trigonometrik Denklemlerin Genel Çözümü
\( k \in Z \) olmak üzere;
\( \cos f(x) = \cos g(x) \Rightarrow f(x) = g(x) + 2k\pi \) veya \( f(x) = -g(x) + 2k\pi \)
\( \sin f(x) = \sin g(x) \Rightarrow f(x) = g(x) + 2k\pi \) veya \( f(x) = \pi \ – \ g(x) + 2k\pi \)
\( \left. \begin{array}{l} \tan f(x) = \tan g(x) \\ \cot f(x) = \cot g(x) \end{array} \right\} \Rightarrow f(x) = g(x) + k\pi \) dir.
Örnek:
\( \cos (2x \ – \ \displaystyle \frac{\pi}{3}) = \cos (\displaystyle \frac{\pi}{4} \ – \ x) \) denkleminin çözüm kümesini bulalım.
\[ \cos (2x \ – \ \displaystyle \frac{\pi}{3}) = \cos (\displaystyle \frac{\pi}{4} \ – \ x) \]
\[ \Rightarrow 2x \ – \ \displaystyle \frac{\pi}{3} = \displaystyle \frac{\pi}{4} \ – \ x + 2k\pi \]
\( \Rightarrow x = \displaystyle \frac{7\pi}{36} + \displaystyle \frac{2k\pi}{3} = \displaystyle \frac{7\pi + 24k\pi}{36} \) veya
\[ 2x \ – \ \displaystyle \frac{\pi}{3} = -(\displaystyle \frac{\pi}{4} \ – \ x) + 2k\pi \]
\( x = \displaystyle \frac{\pi}{12} + 2k\pi = \displaystyle \frac{\pi + 24k\pi}{12} \quad (k \in Z) \) dir.
O halde \( k \in Z \) olmak üzere,
\( Ç = \{ x \mid x = \displaystyle \frac{7\pi + 24k\pi}{36} \quad \text{ veya } \quad x = \displaystyle \frac{\pi + 24k\pi}{12} \} \) dir.
Örnek:
\( \sin (\displaystyle \frac{\pi}{3} \ – \ x) + \cos x = 0 \) denkleminin çözüm kümesini bulalım.
\[ \sin (\displaystyle \frac{\pi}{3} \ – \ x) + \cos x = 0 \]
\[ \Rightarrow \cos x = -\sin (\displaystyle \frac{\pi}{3} \ – \ x) \]
\[ \Rightarrow \cos x = \sin (-\displaystyle \frac{\pi}{3} + x) \]
\( \Rightarrow \sin (\displaystyle \frac{\pi}{2} \ – \ x) = \sin (-\displaystyle \frac{\pi}{3} + x) \) olur.
Buna göre,
\[ -\displaystyle \frac{\pi}{3} + x = \displaystyle \frac{\pi}{2} \ – \ x + 2k\pi \]
\[ \Rightarrow 2x = \displaystyle \frac{\pi}{2} + \displaystyle \frac{\pi}{3} + 2k\pi \]
\( \Rightarrow x = \displaystyle \frac{5\pi}{12} + k\pi \) veya
\[ -\displaystyle \frac{\pi}{3} + x = \pi \ – \ (\displaystyle \frac{\pi}{2} – x) + 2k\pi \]
\[ -\displaystyle \frac{\pi}{3} + x = \pi \ – \ \displaystyle \frac{\pi}{2} + x + 2k\pi \]
\( \Rightarrow -\displaystyle \frac{\pi}{3} \neq \displaystyle \frac{\pi}{2} + 2k\pi \quad (k \in Z) \) dir.
O halde,
\( Ç = \{ x \mid x = \displaystyle \frac{5\pi}{12} + k\pi, \ k \in Z \} \) bulunur.
Örnek:
\( \tan 5x \cdot \tan 3x = 1 \) denkleminin çözüm kümesini bulalım.
\[ \tan 5x \cdot \tan 3x = 1 \Rightarrow \tan 5x = \displaystyle \frac{1}{\tan 3x} \]
\[ \Rightarrow \tan 5x = \cot 3x \]
\[ \Rightarrow \tan 5x = \tan (\displaystyle \frac{\pi}{2} \ – \ 3x) \]
\[ \Rightarrow 5x = \displaystyle \frac{\pi}{2} \ – \ 3x + k\pi \]
\[ \Rightarrow 8x = \displaystyle \frac{\pi}{2} + k\pi \]
\( \Rightarrow x = \displaystyle \frac{\pi}{16} + \displaystyle \frac{k\pi}{8} \quad (k \in Z) \) dir.
O halde,
\( Ç = \{ x \mid x = \displaystyle \frac{\pi}{16} + \displaystyle \frac{k\pi}{8}, \ k \in Z \} \) bulunur.
SORU 64
\( \cos (x + \displaystyle \frac{\pi}{4}) + \sin (\displaystyle \frac{\pi}{4} \ – \ x) = 2 \cos (x + \displaystyle \frac{2\pi}{3}) \) denklemini sağlayan en küçük pozitif x değeri aşağıdakilerden hangisidir?
\[ A) \frac{13\pi}{24} \quad B) \frac{7\pi}{12} \quad C) \frac{5\pi}{8} \quad D) \frac{3\pi}{4} \quad E) \frac{17\pi}{24} \]
Çözüm:
\[ \cos (x + \displaystyle \frac{\pi}{4}) + \sin (\displaystyle \frac{\pi}{4} \ – \ x) = 2 \cos (x + \displaystyle \frac{2\pi}{3}) \]
\[ \Rightarrow \cos (x + \displaystyle \frac{\pi}{4}) + \cos [\displaystyle \frac{\pi}{2} \ – \ (\displaystyle \frac{\pi}{4} – x)] = 2 \cos (x + \displaystyle \frac{2\pi}{3}) \]
\[ \Rightarrow \cos (x + \displaystyle \frac{\pi}{4}) + \cos (\displaystyle \frac{\pi}{4} + x) = 2 \cos (x + \displaystyle \frac{2\pi}{3}) \]
\[ \Rightarrow \cos (x + \displaystyle \frac{\pi}{4}) = \cos (x + \displaystyle \frac{2\pi}{3}) \]
\[ \Rightarrow x + \displaystyle \frac{\pi}{4} = x + \displaystyle \frac{2\pi}{3} + 2k\pi \Rightarrow \displaystyle \frac{\pi}{4} \neq \displaystyle \frac{2\pi}{3} + 2k\pi \]
veya \( x + \displaystyle \frac{\pi}{4} = -(x + \displaystyle \frac{2\pi}{3}) + 2k\pi \)
\[ \Rightarrow 2x = -\displaystyle \frac{\pi}{4} – \displaystyle \frac{2\pi}{3} + 2k\pi \]
\( \Rightarrow x = -\displaystyle \frac{11\pi}{24} + k\pi \) olur.
\( k = 1 \) için \( x = -\displaystyle \frac{11\pi}{24} + \pi = \displaystyle \frac{13\pi}{24} \) tür.
\( \textbf{Cevap : A} \)
SORU 65
\( \displaystyle \frac{\cos x}{\sin 7x} = \cot 3x \) denklemini sağlayan en küçük pozitif x değeri aşağıdakilerden hangisidir?
\[ A) \frac{\pi}{16} \quad B) \frac{\pi}{12} \quad C) \frac{\pi}{10} \quad D) \frac{\pi}{8} \quad E) \frac{\pi}{6} \]
Çözüm:
\[ \displaystyle \frac{\cos x}{\sin 7x} = \cot 3x \Rightarrow \displaystyle \frac{\cos x}{\sin 7x} = \displaystyle \frac{\cos 3x}{\sin 3x} \]
\[ \Rightarrow \sin 3x \cdot \cos x = \sin 7x \cdot \cos 3x \]
\[ \Rightarrow \displaystyle \frac{1}{2} (\sin 4x + \sin 2x) = \displaystyle \frac{1}{2} (\sin 10x + \sin 4x) \]
\[ \Rightarrow \sin 2x = \sin 10x \]
\[ \Rightarrow 10x = 2x + 2k\pi \text{ veya } 10x = \pi – 2x + 2k\pi \]
\[ \Rightarrow x = \displaystyle \frac{k\pi}{4} \qquad x = \displaystyle \frac{(2k + 1)\pi}{12} \]
\( k = 0 \) için \( x = 0 \) veya \( x = \displaystyle \frac{\pi}{12} \) olur.
Buna göre x in en küçük pozitif değeri \( \displaystyle \frac{\pi}{12} \) dir.
\( \textbf{Cevap : B} \)
SORU 66
\( \cos^2 x \ – \ \cos^2 (x \ – \ \displaystyle \frac{\pi}{3}) = 0 \) denkleminin \( [0, \pi) \) aralığındaki köklerinin toplamı nedir?
\[ A) \frac{\pi}{6} \quad B) \frac{\pi}{3} \quad C) \frac{2\pi}{3} \quad D) \frac{4\pi}{3} \quad E) \frac{5\pi}{6} \]
Çözüm:
\[ \cos^2 x \ – \ \cos^2 (x – \displaystyle \frac{\pi}{3}) = 0 \]
\[ \Rightarrow [\cos x – \cos (x \ – \ \displaystyle \frac{\pi}{3})] \cdot [\cos x + \cos (x \ – \ \displaystyle \frac{\pi}{3})] = 0\]
\[ \Rightarrow -2 \sin (x \ – \ \displaystyle \frac{\pi}{6}) \sin \displaystyle \frac{\pi}{6} \cdot 2 \cos (x \ – \ \displaystyle \frac{\pi}{6}) \cos \displaystyle \frac{\pi}{6} = 0 \]
\[ \Rightarrow -2 \sin \displaystyle \frac{\pi}{6} \cos \displaystyle \frac{\pi}{6} \cdot 2 \sin (x \ – \ \displaystyle \frac{\pi}{6}) \cos (x \ – \ \displaystyle \frac{\pi}{6}) = 0\]
\[ \Rightarrow \sin (2x \ – \ \displaystyle \frac{2\pi}{6}) = 0\]
\[ \Rightarrow \sin (2x \ – \ \displaystyle \frac{\pi}{3}) = 0 \]
\[ \Rightarrow \sin (2x \ – \ \displaystyle \frac{\pi}{3}) = \sin 0 = \sin \pi \]
\[ \Rightarrow 2x \ – \ \displaystyle \frac{\pi}{3} = 0 + 2k\pi \quad \text{ veya } \quad 2x – \displaystyle \frac{\pi}{3} = \pi + 2k\pi\]
\[ \Rightarrow x = \displaystyle \frac{\pi}{6} + k\pi \qquad x = \displaystyle \frac{2\pi}{3} + k\pi \]
O halde,
\( k = 0 \) için \( x_1 = \displaystyle \frac{\pi}{6} \) veya \( x_2 = \displaystyle \frac{2\pi}{3} \) olur.
Buna göre,
\( x_1 + x_2 = \displaystyle \frac{\pi}{6} + \displaystyle \frac{2\pi}{3} = \displaystyle \frac{5\pi}{6} \) dır.
\( \textbf{Cevap : E} \)