Number of Solutions

Geometry Level 3

Find the number of values of x x in the interval [ 0 , 3 π ] [0,3\pi] satisfying the equation 2 sin ⁡ 2 x + 5 sin ⁡ x − 3 = 0 2\sin^2{x}+5\sin{x}-3=0 .

6 2 4 8 None of these choices 5

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Sabhrant Sachan
May 12, 2016

Relevant wiki: Trigonometric Equations - Problem Solving - Easy

we have 2 sin ⁡ 2 x + 5 sin ⁡ x − 3 = 0 2\sin^2{x}+5\sin{x}-3=0

⟹ 2 sin ⁡ 2 x + 6 sin ⁡ x − sin ⁡ x − 3 = 0 ⟹ 2 sin ⁡ x ( sin ⁡ x + 3 ) − ( sin ⁡ x + 3 ) = 0 ⟹ sin ⁡ x = − 3 , 1 2 sin ⁡ x = − 3 is rejected , sin ⁡ x = 1 2 ⟹ x = π 6 , 5 π 6 , 13 π 6 , 17 π 6 Total Number of solutions within the range is 4 \implies 2\sin^2{x}+6\sin{x}-\sin{x}-3=0 \\ \implies 2\sin{x}(\sin{x}+3)-(\sin{x}+3)=0 \implies\sin{x}=-3,\dfrac{1}{2}\\ \sin{x}=-3 \text{ is rejected },\sin{x}=\dfrac12\implies x=\dfrac{\pi}{6},\dfrac{5\pi}{6},\dfrac{13\pi}{6},\dfrac{17\pi}{6} \\ \text{Total Number of solutions within the range is 4}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Did the same way in degrees.

Niranjan Khanderia - 5 years, 1 month ago

nice solution +1 ... but not level 4 question , think so , it can be done using also ,

Rudraksh Sisodia - 4 years, 10 months ago

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