Trigonometry! #72

Geometry Level 4

The solution set of ∣ 4 sin ⁡ ( x ) − 1 ∣ < 5 |4\sin (x) - 1| < \sqrt {5} for ∣ x ∣ < π |x|<\pi is

Options:

  1. ( − π , − 4 π 5 ) ∪ ( − π 5 , π 10 ) ∪ ( 9 π 10 , π ) \left(-\pi,-\frac {4\pi}{5} \right)\cup\left(-\frac {\pi}{5} , \frac {\pi}{10} \right) \cup \left( \frac {9\pi}{10} , \pi\right)

  2. ( − 9 π 10 , − π 10 ) ∪ ( 3 π 10 , 7 π 10 ) \left(-\frac {9\pi}{10}, -\frac {\pi}{10}\right) \cup \left(\frac {3\pi}{10} , \frac {7\pi}{10} \right)

  3. ( − π , − 9 π 10 ) ∪ ( − π 10 , 3 π 10 ) ∪ ( 7 π 10 , π ) \left(-\pi,-\frac {9\pi}{10}\right)\cup\left(-\frac {\pi}{10},\frac{3\pi}{10}\right)\cup\left(\frac{7\pi}{10},\pi\right)

  4. ( − 7 π 10 , 9 π 10 ) \left(-\frac {7\pi}{10} , \frac {9\pi}{10}\right)

Note: Enter the serial number of the correct option.

This problem is part of the set Trigonometry .


The answer is 3.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

∣ 4 S i n ( x ) − 1 ∣ < 5 ⟹ 1 − 5 4 < S i n ( x ) < 1 + 5 4 ∴ 0.1 π < x < 0.3 π . T h i s i s a S i n f u c t i o n , s o s o l u t i o n a l s o i n c l u d e s u p p l i m e n t o r y a n g l e s . ∴ S o l u t i o n i s ( 0.1 π , 0.3 π ) U { ( 1 − 0.3 ) π , ( 1 − ( − 0.1 ) π } B u t ∣ x ∣ < π . ∴ { ( 1 − 0.3 ) π , ( 1 − ( − 0.1 ) ) π } c a n b e w r i t t e n a s ( 0.7 π , π ) U ( − π , − 0.9 π ) . S o ( − π , − 9 π 10 ) ∪ ( − π 10 , 3 π 10 ) ∪ ( 7 π 10 , π ) . |4Sin(x)-1|<\sqrt5\\ \implies~\dfrac{1-\sqrt5} 4 <Sin(x)<\dfrac{1+\sqrt5} 4\\ \therefore~\color{#D61F06}{0.1\pi<x<0.3 \pi.}\\ This~is~a~Sin~fuction, ~so ~solution~also ~include~supplimentory~angles.\\ \therefore ~Solution~is ~(0.1\pi ,0.3\pi)~U~\{(1-0.3)\pi,(1-(-0.1)\pi\} \\ But ~~|x|< \pi.\\ \therefore \{(1-0.3)\pi,(1-(-0.1))\pi\} ~can~ be~ written~ as~(0.7\pi,\pi)U( -\pi,-0.9\pi). \\ So~\left(-\pi,-\dfrac {9\pi}{10}\right)\cup\left(-\dfrac {\pi}{10},\dfrac{3\pi}{10}\right)\cup\left(\dfrac{7\pi}{10},\pi\right).

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