Solving Trigonometric equations – ( Qn 1 – 10)

Question No: 1 [ Maximum Mark: 6 ] (Do not use GDC)

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Solve the equation \( \cos ^2 x+\cos x=\sin ^2 x\) where \( 0 \leqslant x \leqslant \pi\)
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Answer : \(x=\pi ; \quad x=\pi / 3 \)

	\( \cos ^2 x+\cos x=\sin ^2 x\)
	Substitute \( \sin ^2 x=1-\cos ^2 x\) \( \begin{aligned} & \cos ^2 x+\cos x=1-\cos ^2 x \\ & 2 \cos ^2 x+\cos x-1=0 \end{aligned}\)	(M1) (A1)
	Substitute \( y=\cos x\) we get,
	\( \begin{aligned} & 2 y^2+y-1=0 \\ & 2 y^2+2 y-1 y-1=0 \\ & 2 y(y+1)-1(y+1)=0 \\ & (y+1)(2 y-1)=0 \\ & y+1=0 ; \quad 2 y-1=0 \\ & y=-1 \quad y=\frac{1}{2} \end{aligned}\)	(M1) (A1)
	Replace \( y=\cos x\), we get,
	\(\cos x=-1\Rightarrow \) \( x=\pi\)
	\(\cos x=\frac{1}{2} \Rightarrow\) ; \(x=\pi / 3 \)
	Therefore \(x=\pi ; \quad x=\pi / 3 \)	( A1 ) ( A1)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

Solve the equation \( 3 \cos ^2 x-3=\sin ^{2} x-1\) where \( 0<x<2 \pi\)

Answer : \(\begin{aligned}
& x=\frac{\pi}{6} , \quad11 \frac{\pi}{6}
,\quad5 \frac{\pi}{6} ,\quad 7 \frac{\pi}{6}
\end{aligned} \)

	\( 3 \cos ^2 x-3=\sin ^2 x-1\)
	\(3 \cos ^2 x-\sin ^2 x-3+1=0 \)	( M1 )
	Substitute \( \sin ^2 x=1-\cos ^2 x\) we get \(3 \cos ^2 x-1+\cos ^2 x-2=0 \) \( 4 \cos ^2 x-3=0\) \(\cos ^2 x=\frac{3}{4} \) \(\cos x= \pm \frac{\sqrt{3}}{2} \) \( \cos x=+\frac{\sqrt{3}}{2} \Rightarrow x=\frac{\pi}{6} ; 11 \frac{\pi}{6}\) \( \cos x=-\frac{\sqrt{3}}{2} \Longrightarrow x=5 \frac{\pi}{6} ; 7 \frac{\pi}{6}\)	( A1 ) ( M1) ( A1) ( A1) ( A1)
	Therefore the solutions are \(\begin{aligned} & x=\frac{\pi}{6} , \quad11 \frac{\pi}{6} ,\quad5 \frac{\pi}{6} ,\quad 7 \frac{\pi}{6} \end{aligned} \)	( A1)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

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Consider the quadratic equation \( x^2-\sqrt{8} \cos \theta \cdot x+3 \cos \theta=1 \text {. }\) where\( 0 \leq \theta \leq 2 \pi\) has equal roots , find the values of \( \theta\)
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Answer : \( \theta=\frac{\pi}{3}, \quad\frac{5 \pi}{3},\quad0,\quad2 \pi\)

\( x^2-\sqrt{8} \cos \theta \cdot x+3 \cos \theta=1 \text {. }\)

\( x^2-\sqrt{8} \cos \theta \cdot x+3 \cos \theta-1 =0 \text {. }\)

Since the equation has two equal roots, the discernment \( b^2-4 a c=0\)

Therefore \(8 \cos ^2 \theta-4(3 \cos \theta-1)=0\)

divide by 4 we get \( 2 \cos ^2 \theta-3 \cos \theta+1=0 .\)

\((2 \cos \theta-1)(\cos \theta-1)=0 \)

\( \cos \theta=\frac{1}{2}\Rightarrow \theta=\frac{\pi}{3}, \frac{5 \pi}{3}\)

\( \cos \theta=1 . \Longrightarrow \theta=0,2 \pi .\)

(M1)

(A1)

(M1)

(A1)

(A1

(A1)

Therefore the solutions are \( \theta=\frac{\pi}{3}, \quad\frac{5 \pi}{3},\quad0,\quad2 \pi\)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Consider the equation \( 2 \sin ^2 x+\sin ^2 2 x=2\) , find the value of \( \cos x\) where \( 0 \leq x \leq \pi / 2\) ;
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Answer : \(x=\frac{\pi}{2}, \quad \frac{\pi}{4} \)

\(2 \sin ^2 x+\sin ^2 2 x=2 \)

\(2 \sin ^2 x+(2 \sin x \cos x)^2=2 \)

\( 2 \sin ^2 x+4 \sin ^2 x \cos ^2 x=2\)

\( \sin ^2 x+2 \sin ^2 x \cos ^2 x=1\)

\(\left(1-\sin ^2 x\right)-2 \sin ^2 x \cos ^2 x=0 \)

\(\cos ^2 x-2 \sin ^2 x \cos ^2 x=0 \)

\(\Rightarrow \cos ^2 x\left(1-2 \sin ^2 x\right)=0 \)

\( \cos x=0 \Rightarrow x=\pi / 2\)

\( \begin{aligned}
& 1-2 \sin ^2 x=0 \\
& 2 \sin ^2 x=1 \\
& \sin ^2 x=\frac{1}{2} \\
& \sin x= \pm 1 / \sqrt{2}
\end{aligned}\)

\( \sin x=\frac{1}{\sqrt{2}} \Rightarrow x=\pi / 4\)

\(\sin x=\frac{-1}{\sqrt{2}} \text { No soiutrom from } 0 \leqslant x \leqslant \pi / 2 \)

Therefore the solutions are \(x=\frac{\pi}{2}, \frac{\pi}{4} \)

(M1)

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Question No: 5 [ Maximum Mark: 4 ] (Do not use GDC)

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Consider the equation \( \sin x=\frac{3}{4}\) where \( 0 \leq x \leq \pi / 2\), find the values of \( \sin x\)
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Answer : \(\sin x=-\frac{1}{3} \text { or } \sin x=1 \text {. } \)

\(3 \sin x-\frac{1}{\sin x}=2, \)

\( 3 \sin ^2 x-2 \sin x-1=0\)

(M1)

\( \left(3 \sin x^{\circ}+1\right)\left(\sin x^{\circ}-1\right)=0\)

(A1)

\(\sin x=-\frac{1}{3} \quad \text { or } \sin x=1 \text {. } \)

(A1) (A1)

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Question No: 6 [ Maximum Mark: 7 ] (Do not use GDC)

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Given \( \cos 2 x-\cos x=0\) where \( 0 \leq x \leq 2\pi \) ;
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Answer : \( \)

	a \( \)	(M1)
	a \( \)
	a \( \)
	a \( \)
	a \( \)

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Question No: 7 [ Maximum Mark: 7 ] (Do not use GDC)

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Given \( 5 \cos ^2 x=4-3 \sin ^2 x\) where \( \pi \leq x \leq \pi \) ;
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Answer : \(-\frac{3\pi}{4}, -\frac{\pi}{4},\frac{\pi}{4}, \frac{3 \pi}{4}\)

	a \( \)	(M1)
	a \( \)
	a \( \)
	a \( \)
	a \( \)

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Question No: 8 [ Maximum Mark: 7 ] (Do not use GDC)

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Given \( \sin x=\frac{3}{4}\) where \( 0 \leq x \leq \pi / 2\) ;

(a)	Find \(\cos x \)		[3 marks]
(b)	Find \(\cos x \)		[3 marks]

Answer : \( \)

	a \( \)	(M1)
	a \( \)
	a \( \)
	a \( \)
	a \( \)

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Question No: 9 [ Maximum Mark: 7 ] (Do not use GDC)

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Given \( \sin x=\frac{3}{4}\) where \( 0 \leq x \leq \pi / 2\) ;

(a)	Find \(\cos x \)		[3 marks]
(b)	Find \(\cos x \)		[3 marks]

Answer : \( \)

	a \( \)	(M1)
	a \( \)
	a \( \)
	a \( \)
	a \( \)

Question No: 1 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 4 ] (Do not use GDC)

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Question No: 6 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 7 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 8 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 9 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 10 [ Maximum Mark: 7 ] (Do not use GDC)

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Solving Trigonometric equations – ( Qn 1 – 10)

Question No: 1 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 2 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 3 [ Maximum Mark: 6 ] (Do not use GDC)

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Question No: 4 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 5 [ Maximum Mark: 4 ] (Do not use GDC)

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Question No: 6 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 7 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 8 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 9 [ Maximum Mark: 7 ] (Do not use GDC)

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Question No: 10 [ Maximum Mark: 7 ] (Do not use GDC)

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General

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