Let ${S_1} = \{ x \in [0,12\pi ]:{\sin ^5}x + {\cos ^5}x = 1\}$
and ${S_2} = \{ x \in [0,8\pi ]:{\sin ^7}x + {\cos ^7}x = 1\}$
Then $n({S_1}) - n({S_2})$ is equal to ______________.
Answer (integer)
4
Solution
<p>Given,</p>
<p>$${S_1} = \left\{ {x \in \left[ {0,12\pi } \right],\,{{\sin }^5}x + {{\cos }^5}x = 1} \right\}$$</p>
<p>$${S_2} = \left\{ {x \in \left[ {0,8\pi } \right],\,{{\sin }^7}x + {{\cos }^7}x = 1} \right\}$$</p>
<p>(1) ${\sin ^5}x + {\cos ^5}x = 1$</p>
<p>This satisfies when $\sin x = 1$ and $\cos x = 0$</p>
<p>$\therefore$ $$x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2},{{17\pi } \over 2},{{21\pi } \over 2}$$</p>
<p>It also satisfies when $\sin x = 0$ and $\cos x = 1$</p>
<p>$\therefore$ $x = 0,\,2\pi ,4\pi ,6\pi ,8\pi ,10\pi ,12\pi$</p>
<p>$\therefore$ Accepted values of x in $\left[ {0,12\pi } \right]$ is = 13</p>
<p>$\therefore$ $n({S_1}) = 13$</p>
<p>(2) ${\sin ^7}x + {\cos ^7}x = 1$</p>
<p>This satisfies when $\sin x = 1$ and $\cos x = 0$</p>
<p>For $x \in \left[ {0,\,8\pi } \right]$, possible values</p>
<p>$x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$</p>
<p>It also satisfies when $\sin x = 0$ and $\cos x = 1$</p>
<p>$x \in \left[ {0,8\pi } \right]$, possible values</p>
<p>$x = 0,2\pi ,4\pi ,6\pi ,8\pi$</p>
<p>$\therefore$ Total accepted values of x in $\left[ {0,8\pi } \right]$ is = 9</p>
<p>$\therefore$ $n({S_2}) = 9$</p>
<p>$\therefore$ $n({S_1}) - n({S_2}) = 13 - 9 = 4$</p>
About this question
Subject: Mathematics · Chapter: Trigonometric Functions · Topic: Trigonometric Ratios and Identities
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