The number of solutions of the equation (4-sqrt(3))sin x - 2sqrt(3)cos^2 x = -4/(1+sqrt(3)), x in [-2pi, 5pi/2] is

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Maths Q13 – JEE-MAIN 2025

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Home Maths Trigonometric FunctionsQuestion #493

The number of solutions of the equation

(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}

x \in [- 2 π, \frac{5 π}{2}]

Clue 1 of 3: Simplify the equation

Begin by simplifying the constant term on the right-hand side and then express $\cos^{2} x$ in terms of $\sin^{2} x$ to form a quadratic equation in $\sin x$ .

🧠 Full Solution Path

Step 1: Simplify the equation to a quadratic in

\sin x

✦ Active

The given equation is $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}$ . First, simplify the RHS:

- \frac{4}{1 + \sqrt{3}} = - \frac{4 (\sqrt{3} - 1)}{(\sqrt{3} + 1) (\sqrt{3} - 1)} = - \frac{4 (\sqrt{3} - 1)}{2} = - 2 (\sqrt{3} - 1) = 2 - 2 \sqrt{3}

Substitute $\cos^{2} x = 1 - \sin^{2} x$ into the equation:

(4 - \sqrt{3}) \sin x - 2 \sqrt{3} (1 - \sin^{2} x) = 2 - 2 \sqrt{3}

Rearranging and simplifying, we get a quadratic equation in $\sin x$ :

2 \sqrt{3} \sin^{2} x + (4 - \sqrt{3}) \sin x - 2 = 0

Step 2: Solve the quadratic equation for

\sin x

○ Expand

Let $y = \sin x$ . The quadratic equation is $2 \sqrt{3} y^{2} + (4 - \sqrt{3}) y - 2 = 0$ . Using the quadratic formula $y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

The discriminant $D = (4 - \sqrt{3})^{2} - 4 (2 \sqrt{3}) (- 2) = 16 - 8 \sqrt{3} + 3 + 16 \sqrt{3} = 19 + 8 \sqrt{3}$ . This can be written as $(4 + \sqrt{3})^{2}$ .

y = \frac{- (4 - \sqrt{3}) \pm \sqrt{(4 + \sqrt{3})^{2}}}{2 (2 \sqrt{3})} = \frac{- 4 + \sqrt{3} \pm (4 + \sqrt{3})}{4 \sqrt{3}}

This yields two possible values for $y$ :

y_{1} = \frac{- 4 + \sqrt{3} + 4 + \sqrt{3}}{4 \sqrt{3}} = \frac{2 \sqrt{3}}{4 \sqrt{3}} = \frac{1}{2}

y_{2} = \frac{- 4 + \sqrt{3} - (4 + \sqrt{3})}{4 \sqrt{3}} = \frac{- 8}{4 \sqrt{3}} = - \frac{2}{\sqrt{3}} = - \frac{2 \sqrt{3}}{3}

Since $\sin x$ must be in $[- 1, 1]$ , $y_{2} = - \frac{2 \sqrt{3}}{3} \approx - 1.154$ is rejected. Thus, we only consider $\sin x = \frac{1}{2}$ .

💡 Teacher's Secret Hint

Remember to check the domain of $\sin x$ after solving the quadratic equation.

Step 3: Find the number of solutions in the given interval○ Expand

We need to find solutions for $\sin x = \frac{1}{2}$ in the interval $x \in [- 2 π, \frac{5 π}{2}]$ . This interval can be written as $[- \frac{12 π}{6}, \frac{15 π}{6}]$ .

The general solutions for $\sin x = \frac{1}{2}$ are $x = n π + (- 1)^{n} \frac{π}{6}$ , where $n \in Z$ . Let's list the solutions within the given interval:

For $n = 0 : x = \frac{π}{6}$ (valid).

For $n = 1 : x = π - \frac{π}{6} = \frac{5 π}{6}$ (valid).

For $n = 2 : x = 2 π + \frac{π}{6} = \frac{13 π}{6}$ (valid, as $\frac{13 π}{6} < \frac{15 π}{6}$ ). The next positive solution $x = 3 π - \frac{π}{6} = \frac{17 π}{6}$ is outside the interval.

For $n = - 1 : x = - π - \frac{π}{6} = - \frac{7 π}{6}$ (valid, as $- \frac{12 π}{6} < - \frac{7 π}{6}$ ).

For $n = - 2 : x = - 2 π + \frac{π}{6} = - \frac{11 π}{6}$ (valid, as $- \frac{12 π}{6} < - \frac{11 π}{6}$ ). The next negative solution $x = - 3 π - \frac{π}{6} = - \frac{19 π}{6}$ is outside the interval.

Counting these, there are 5 distinct solutions in the given interval.

💡 Teacher's Secret Hint

Carefully check the interval boundaries for each potential solution.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Simplify the equation

🧠 Full Solution Path