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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The sum of all the values of x between 0 and 4π which satisfy the...

The sum of all the values of xx between 00 and 4π4\pi which satisfy the equation sin⁡x8cos2x=1\sin x\sqrt{8co{s}^{2}x} = 1 is kπk\pi, then the value of k5\dfrac{k}{5} is equal to

Correct answer

2.4

Step-by-step solution

Given equation is sin⁡xcos2x=122\sin x\sqrt{co{s}^{2}x} = \dfrac{1}{2\sqrt{2}}
⇒sinx∣cos⁡x∣=122⇒ sin x\left|\cos x\right| = \dfrac{1}{2\sqrt{2}}
Case (I): When x>0x > 0 i.e. xx lies in 1st{1}^{st} or 4th{4}^{th} quadrant.
sinxcosx=122⇒sin⁡2x=12sin xcos x = \dfrac{1}{2\sqrt{2}} ⇒ \sin 2x = \dfrac{1}{\sqrt{2}}
⇒2x=π4,3π4,9π4,11π4,17π4,19π4⇒ 2x = \dfrac{\pi}{4}, \dfrac{3\pi}{4}, \dfrac{9\pi}{4}, \dfrac{11\pi}{4}, \dfrac{17\pi}{4}, \dfrac{19\pi}{4}
⇒x=π8,3π8,9π8,11π8,17π8,19π8⇒ x = \dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{9\pi}{8}, \dfrac{11\pi}{8}, \dfrac{17\pi}{8}, \dfrac{19\pi}{8}
As xx lies between 00 and 4π,4\pi, cosx>0⇒x=π8,3π8,17π8,19π8cos x > 0 ⇒ x = \dfrac{\pi}{8}, \dfrac{3\pi}{8}, \dfrac{17\pi}{8}, \dfrac{19\pi}{8}
Case (II): When cosx<0cos x < 0 i.e., xx lies in 2nd{2}^{nd} or 3rd{3}^{rd} quadrant.
sin⁡2x=−12⇒2x=5π4,7π4,13π4,15π4,21π4,23π4\sin 2x = \dfrac{- 1}{\sqrt{2}} ⇒ 2x = \dfrac{5\pi}{4}, \dfrac{7\pi}{4}, \dfrac{13\pi}{4}, \dfrac{15\pi}{4}, \dfrac{21\pi}{4}, \dfrac{23\pi}{4}
⇒x=5π8,7π8,13π8,15π8,21π8,23π8⇒ x = \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{13\pi}{8}, \dfrac{15\pi}{8}, \dfrac{21\pi}{8}, \dfrac{23\pi}{8}
As xx lies between 00 and 4π,4\pi, cosx<0⇒x=5π8,7π8,21π8,23π8cos x < 0 ⇒ x = \dfrac{5\pi}{8}, \dfrac{7\pi}{8}, \dfrac{21\pi}{8}, \dfrac{23\pi}{8}
Sum of values of x=π8(1+3+17+19+5+7+21+23)=96π8=12πx = \dfrac{\pi}{8}\left(1 + 3 + 17 + 19 + 5 + 7 + 21 + 23\right) = \dfrac{96\pi}{8} = 12\pi
⇒k=12⇒k5=2.4⇒ k = 12 ⇒ \dfrac{k}{5} = 2 . 4
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