JEE Main Previous Year Trigonometry Questions and Solutions

Q 1 :
Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

$\frac{6 - \sqrt{6}}{(3 \sqrt{6} - 2)}$
$\frac{4}{(3 \sqrt{6} - 2)}$
$\frac{4}{(3 \sqrt{6} + 2)}$
$\frac{6 + \sqrt{6}}{(3 \sqrt{6} + 2)}$

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(2)

We have, $4 \cos θ - 3 \sin θ = 1$

$\Rightarrow 4 \cos θ - 1 = 3 \sin θ \Rightarrow {(4 \cos θ - 1)}^{2} = {(3 \sin θ)}^{2}$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 \sin^{2} θ$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 (1 - \cos^{2} θ)$

$\Rightarrow 25 \cos^{2} θ - 8 \cos θ - 8 = 0$

$\Rightarrow \cos θ = \frac{8 \pm \sqrt{64 + 800}}{50}$

$= \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12 \sqrt{6}}{50} = \frac{4 \pm 6 \sqrt{6}}{25}$

Since, $θ \in [0, π / 4]$ so $\cos θ = \frac{4 + 6 \sqrt{6}}{25}$

$= \frac{(4 - 6 \sqrt{6}) (4 + 6 \sqrt{6})}{25 (4 - 6 \sqrt{6})} = \frac{- 200}{25 \times 2 (2 - 3 \sqrt{6})} = \frac{4}{3 \sqrt{6} - 2}$

Q 2 :
Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

$18 π$
$9 π$
$6 π$
$15 π$

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(3)

$| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow$ $| \cos θ \cos (\frac{π}{3} - θ) \cos (\frac{π}{3} + θ) |$ $\leq \frac{1}{8}$

$\Rightarrow \frac{1}{4} | \cos 3 θ | \leq \frac{1}{8} \Rightarrow | \cos 3 θ | \leq \frac{1}{2} \Rightarrow \frac{- 1}{2} \leq \cos 3 θ \leq \frac{1}{2}$

$cos 3 θ is max if \cos 3 θ = \frac{1}{2}$

$\Rightarrow 3 θ = 2 n π \pm \frac{π}{3} \Rightarrow θ = \frac{2 n π}{3} \pm \frac{π}{9}$

when $θ \in [0, 2 π]$

$∴$ $θ = \frac{π}{9}, \frac{5 π}{9}, \frac{7 π}{9}, \frac{11 π}{9}, \frac{13 π}{9}, \frac{17 π}{9}$

$Hence, \sum θ_{i} = 6 π$

Q 3 :
The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

0
3
1
2

1

2

3

4

(1)

We have, $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0$

$\Rightarrow 4 (1 - \cos^{2} x) - 4 \cos^{3} x + 9 - 4 \cos x = 0$

$\Rightarrow 4 \cos^{3} x + 4 \cos^{2} x + 4 \cos x = 13$

Ist clear that L.H.S. $\leq 12$ and never 13

$∴$ Solution does not exists.

Q 4 :
If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

$\frac{\sqrt{10} - 10}{12}$
$\frac{10 - \sqrt{10}}{6}$
$\frac{\sqrt{10} - 10}{6}$
$\frac{10 - \sqrt{10}}{12}$

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4

(1)

We have, $4 \cos θ + 5 \sin θ = 1$

$\Rightarrow 4 (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + 5 (\frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) = 1$

$\Rightarrow 5 \tan^{2} (θ / 2) - 10 \tan (θ / 2) - 3 = 0$

$\Rightarrow \tan (θ / 2) = \frac{10 \pm \sqrt{100 - 4 (5) (- 3)}}{2 \times 5} = \frac{10 \pm \sqrt{160}}{10} = \frac{5 \pm \sqrt{40}}{5}$

$∵$ $α \in (\frac{- π}{2}, \frac{π}{2})$ So, $\frac{α}{2} \in (\frac{- π}{4}, \frac{π}{4})$

$\Rightarrow \tan (\frac{α}{2}) \in (- 1, 1)$ $∴$ $\tan \frac{α}{2} = \frac{5 - \sqrt{40}}{5}$

$Hence, \tan α = \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}} = \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - {(1 - \frac{\sqrt{40}}{5})}^{2}}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - (1 + \frac{40}{25} - \frac{2 \sqrt{40}}{5})}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{\frac{2 \sqrt{40}}{5} - \frac{8}{5}} = \frac{5 - \sqrt{40}}{\sqrt{40} - 4} \times \frac{\sqrt{40} + 4}{\sqrt{40} + 4}$

$= \frac{5 \sqrt{40} + 20 - 40 - 4 \sqrt{40}}{40 - 16} = \frac{\sqrt{40} - 20}{24} = \frac{\sqrt{10} - 10}{12}$

Q 5 :
The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

0
- 1
1
3

1

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3

4

(2)

We have, $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{\cos 2 x (3 + \cos^{2} 2 x)}{(\cos^{2} x - \sin^{2} x) ((\cos^{2} x + \sin^{2} x)^{2} - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{(3 + \cos^{2} 2 x)}{(1 - \cos^{2} x \sin^{2} x)} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x - \sin^{2} x)}^{2}}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{3 + {(\cos^{2} x + \sin^{2} x)}^{2} - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow \frac{4 - 4 \cos^{2} x \sin^{2} x}{1 - \cos^{2} x \sin^{2} x} = x^{3} - x^{2} + 6$

$\Rightarrow 4 = x^{3} - x^{2} + 6 \Rightarrow x^{3} - x^{2} + 2 = 0$

$\Rightarrow (x + 1) (x^{2} - 2 x + 2) = 0$

$\Rightarrow x = - 1 is the only real solution.$

Q 6 :
For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

$- \frac{2}{3}$
$\frac{2}{3}$
$5$
$- 5$

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4

(4)

We have, $3 \sin (α + β) = 2 \sin (α - β)$

$\Rightarrow \frac{\sin (α + β)}{\sin (α - β)} = \frac{2}{3} \Rightarrow \frac{\sin (α + β) + \sin (α - β)}{\sin (α + β) - \sin (α - β)} = \frac{5}{- 1}$ (Using componendo and dividendo)

$\Rightarrow \frac{2 \sin α \cos β}{2 \cos α \sin β} = - 5 \Rightarrow \frac{\tan α}{\tan β} = - 5$

$\Rightarrow \tan α = - 5 \tan β$ $∴$ $k = - 5$

Q 7 :
The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

0
2
1
more than 2

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(1)

We have, $e^{\sin x} - 2 e^{- \sin x} = 2$ ...(i)

Let $e^{\sin x} = t$

$∴$ $(i) becomes$ $t - \frac{2}{t} = 2$

$\Rightarrow t^{2} - 2 = 2 t \Rightarrow t^{2} - 2 t - 2 = 0$

$\Rightarrow t = \frac{2 \pm \sqrt{4 - 4 (1) (- 2)}}{2} \Rightarrow t = \frac{2 \pm \sqrt{12}}{2} = 1 \pm \sqrt{3}$

Now, $e^{\sin x} = 1 + \sqrt{3} and e^{\sin x} = 1 - \sqrt{3} (Not possible)$

$\Rightarrow \sin x = \ln (1 + \sqrt{3}) > 1 (Not possible)$

$∴$ $There is no solution.$

Q 8 :
Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

0%

(4)

For real roots $D \geq 0$

$\Rightarrow {(\sin 2 θ)}^{2} - 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ) \geq 0$

$\Rightarrow \sin^{2} 2 θ \geq 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ)$

Put $\sin^{2} 2 θ = t$

$\Rightarrow t \geq 4 (1 - \frac{t}{2}) (1 - \frac{3 t}{4}) \Rightarrow 2 t \geq (2 - t) (4 - 3 t)$

$\Rightarrow 3 t^{2} - 12 t + 8 \leq 0$ $\Rightarrow t^{2} - 4 t + \frac{8}{3} \leq 0$

$\Rightarrow {(t - 2)}^{2} + \frac{8}{3} - 4 \leq 0$ $\Rightarrow {(t - 2)}^{2} \leq \frac{4}{3}$

$\Rightarrow - \frac{2}{\sqrt{3}} \leq t - 2 \leq \frac{2}{\sqrt{3}}$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 2 + \frac{2}{\sqrt{3}}$

$∵$ $t \in [0, 1]$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 1$

Since, $α$ and $β$ be the smallest and largest elements of set $S$

$∴$ $α = 2 - \frac{2}{\sqrt{3}}, β = 1$

$Hence, 3 [{(α - 2)}^{2} + {(β - 1)}^{2}] = 3 [{(2 - \frac{2}{\sqrt{3}} - 2)}^{2} + {(1 - 1)}^{2}]$ $= 3 [\frac{4}{3} + 0] = 4 .$

Q 9 :
Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]

0%

(48)

$\cos 2 x + a \sin x = 2 a - 7$

$1 - 2 \sin^{2} x + a \sin x = 2 a - 7$

Let $t = \sin x$ $∴$ $2 t^{2} - a t + 2 a - 8 = 0$

$\Rightarrow (t - 2) (2 t + 4 - a) = 0 \Rightarrow t = 2 or t = \frac{a - 4}{2}$

$∴$ $\sin x = 2 (not possible) or 2 \sin x = a - 4$

$\Rightarrow a = 2 \sin x + 4 \in [2, 6]$ $[∵$ $\sin x \in [- 1, 1]]$

$So, p = 2, q = 6$

Now, $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °$

$= (\tan 9 ° + \cot 9 °) - (\tan 27 ° + \cot 27 °)$

$= 2 c o s e c 18 ° - 2 c o s e c 54 °$ $[∵$ $\tan θ + \cot θ = 2 c o s e c 2 θ]$

$= 2 [\frac{4}{\sqrt{5} - 1} - \frac{4}{\sqrt{5} + 1}] = 4 \Rightarrow r = 4$

$∴$ $p q r = 2 \times 6 \times 4 = 48$

Topic Question Set

Q 1 :
Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

Q 2 :
Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

Q 3 :
The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

Q 4 :
If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

Q 5 :
The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

Q 6 :
For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

Q 7 :
The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

Q 8 :
Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

0%

Q 9 :
Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]

0%

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Topic Question Set

Q 1 : Suppose θ∈[0,π4] is a solution of 4cosθ-3sinθ=1. Then cosθ is equal to : [2024]

Q 2 : Let |cosθcos(60-θ)cos(60+θ)|≤18,θ∈[0,2π]. Then, the sum of all θ∈[0,2π], where cos3θ attains its maximum value, is: [2024]

Q 3 : The number of solutions of the equation 4sin2x-4cos3x+9-4cosx=0; x∈[-2π,2π] is : [2024]

Q 4 : If α,-π2<α<π2 is the solution of 4cosθ+5sinθ=1, then the value of tanα is [2024]

Q 5 : The sum of the solutions x∈R of the equation 3cos2x+cos32xcos6x-sin6x=x3-x2+6 is [2024]

Q 6 : For α,β∈(0,π2), let 3sin(α+β)=2sin(α-β) and a real number k be such that tanα=k tanβ. Then, the value of k is equal to [2024]

Q 7 : The number of solutions of the equation esinx-2e-sinx=2, is [2024]

Q 8 : Let S={sin22θ:(sin4θ+cos4θ)x2+(sin2θ)x+(sin6θ+cos6θ)=0 has real roots}. If α and β be the smallest and largest elements of the set S, respectively, then 3((α-2)2+(β-1)2) equals _________ . [2024]

0%

Q 9 : Let the set of all a∈R such that the equation cos2x+asinx=2a-7 has a solution be [p,q] and r=tan9°-tan27°-1cot63°+tan81°, then pqr is equal to _____ . [2024]

0%

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Q 1 :
Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

Q 2 :
Let $| \cos θ \cos (60 - θ) \cos (60 + θ) |$ $\leq \frac{1}{8},$ $θ \in [0, 2 π] .$ Then, the sum of all $θ \in [0, 2 π],$ where $\cos 3 θ$ attains its maximum value, is: [2024]

Q 3 :
The number of solutions of the equation $4 \sin^{2} x - 4 \cos^{3} x + 9 - 4 \cos x = 0;$ $x \in [- 2 π, 2 π]$ is : [2024]

Q 4 :
If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

Q 5 :
The sum of the solutions $x \in R$ of the equation $\frac{3 \cos 2 x + \cos^{3} 2 x}{\cos^{6} x - \sin^{6} x} = x^{3} - x^{2} + 6$ is [2024]

Q 6 :
For $α, β \in (0, \frac{π}{2}),$ let $3 \sin (α + β) = 2 \sin (α - β)$ and a real number $k$ be such that $\tan α = k$ $\tan β .$ Then, the value of $k$ is equal to [2024]

Q 7 :
The number of solutions of the equation $e^{\sin x} - 2 e^{- \sin x} = 2,$ is [2024]

Q 8 :
Let $S = {\sin^{2} 2 θ : (\sin^{4} θ + \cos^{4} θ) x^{2} + (\sin 2 θ) x + (\sin^{6} θ + \cos^{6} θ) = 0 has real roots} .$ If $α$ and $β$ be the smallest and largest elements of the set $S,$ respectively, then $3 ({(α - 2)}^{2} + {(β - 1)}^{2})$ equals _________ . [2024]

Q 9 :
Let the set of all $a \in R$ such that the equation $\cos 2 x + a \sin x = 2 a - 7$ has a solution be $[p, q]$ and $r = \tan 9 ° - \tan 27 ° - \frac{1}{\cot 63 °} + \tan 81 °,$ then $p q r$ is equal to _____ . [2024]