The general value of θ satisfying sinθ =-1/2 and tanθ =1√(3) is

Question 6 of 14 · Trigonometric Equations

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NIMCET 2021

The general value of $θ$ satisfying $s in θ = - \frac{1}{2}$ and $t an θ = \frac{1}{3}$ is

Answer (Detailed Solution Below)View answer ▾

Answer: Option C — $2 nπ + \frac{7 π}{6}, n \in I$

Detailed Solution

$s in θ < 0$ and $t an θ > 0$ put $θ$ in the third quadrant.

In $[0, 2 π)$ the angle is $π + \frac{π}{6} = \frac{7 π}{6}$ : $s in \frac{7 π}{6} = - \frac{1}{2}$ and $t an \frac{7 π}{6} = \frac{1}{3}$ .

Both conditions together repeat only every $2 π$ , so $θ = 2 nπ + \frac{7 π}{6}$ .

✓ Check

n = 1

gives

θ = \frac{19 π}{6}

, which is

570^{\circ}

, i.e.

210^{\circ}

again:

s in = - \frac{1}{2}

t an = \frac{1}{3}

⚠ Common trap

(A)

nπ + \frac{π}{6}

solves only

t an θ = \frac{1}{3}

; for even

n

it gives

s in θ = + \frac{1}{2}

Quick tip

$sin < 0$ and $tan > 0 \Rightarrow$ third quadrant: $θ = 2 nπ + \frac{7 π}{6}$

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Trigonometric Equations — Preparation

This question tests your understanding of Trigonometric Equations, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A trigonometric equation is an equation involving trigonometric functions of an unknown angle, such as 2 sin x = 1. This topic covers finding principal solutions (usually between 0 and 2 pi) and general solutions, which include all possible angles. The key general solutions are: sin x = sin a gives x = n pi + (-1)^n a, cos x = cos a gives x = 2n pi +/- a, and tan x = tan a gives x = n pi + a, where n is any integer. Exam questions involve reducing an equation to one of these standard forms, often by factorising or using identities, and then counting the number of solutions in a given interval. Always check that solutions do not make any term in the original equation undefined. Practise the trigonometric equations questions below, each with a step-by-step solution.

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