SATHEE: Trigonometrical Equations 1 Question 22

Trigonometrical Equations 1 Question 22

22. The number of distinct solutions of the equation $\frac{5}{4} \cos^{2} 2 x + \cos^{4} x + \sin^{4} x + \cos^{6} x + \sin^{6} x = 2$ in the interval $[0, 2 π]$ is

(2015 Adv.)

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Answer:

Correct Answer: 22. (8)

Solution:

Here, $\frac{5}{4} \cos^{2} 2 x + (\cos^{4} x + \sin^{4} x) + (\cos^{6} x + \sin^{6} x) = 2$

$\Rightarrow \frac{5}{4} \cot 2 x + [{(\cos^{2} x + \sin^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x]$

$+ (\cos^{2} x + \sin^{2} x) [{(\cos^{2} x + \sin^{2} x)}^{2} - 3 \sin^{2} x \cos^{2} x] = 2$

$\Rightarrow \frac{5}{4} \cos^{2} 2 x + (1 - 2 \sin^{2} x \cos^{2} x) + (1 - 3 \cos^{2} x \sin^{2} x) = 2$

$\Rightarrow \frac{5}{4} \cos^{2} 2 x - 5 \sin^{2} x \cos^{2} x = 0$

$\begin{array}{lc} \Rightarrow & \frac{5}{4} \cos^{2} 2 x - \frac{5}{4} \sin^{2} 2 x = 0 \\ \Rightarrow & \frac{5}{4} \cos^{2} 2 x - \frac{5}{4} + \frac{5}{4} \cos^{2} 2 x = 0 \\ \Rightarrow & \frac{5}{2} \cos^{2} 2 x = \frac{5}{4} \Rightarrow \cos^{2} 2 x = \frac{1}{2} \\ \Rightarrow & 2 \cos^{2} 2 x = 1 \\ \Rightarrow & 1 + \cos 4 x = 1 \\ \Rightarrow & \cos 4 x = 0 as 0 \leq x \leq 2 π \\ ∴ & 4 x = \frac{π}{2}, \frac{3 π}{2}, \frac{5 π}{2}, \frac{7 π}{2}, \frac{9 π}{2}, \frac{11 π}{2}, \frac{13 π}{2}, \frac{15 π}{2} \\ as & 0 \leq 4 x \leq 8 π \\ \Rightarrow & x = \frac{π}{8}, \frac{3 π}{8}, \frac{5 π}{8}, \frac{7 π}{8}, \frac{9 π}{8}, \frac{11 π}{8}, \frac{13 π}{8}, \frac{15 π}{8} \end{array}$

Hence, the total number of solutions is 8 .

Trigonometrical Equations 1 Question 22

22. The number of distinct solutions of the equation $\frac{5}{4} \cos^{2} 2 x + \cos^{4} x + \sin^{4} x + \cos^{6} x + \sin^{6} x = 2$ in the interval $[0, 2 π]$ is

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Trigonometrical Equations 1 Question 22

22. The number of distinct solutions of the equation 54cos2⁡2x+cos4⁡x+sin4⁡x+cos6⁡x+sin6⁡x=2 in the interval [0,2π] is

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22. The number of distinct solutions of the equation $\frac{5}{4} \cos^{2} 2 x + \cos^{4} x + \sin^{4} x + \cos^{6} x + \sin^{6} x = 2$ in the interval $[0, 2 π]$ is