The number of values of - in the interval -2- -2 such that - n-5 f orn -0- - 1- - 2 and tan- 5 - as well as sin 2 -cos 4 - is

Tan θ =cot 5 θ ⇒ cos 6 θ =0 ⇒ 4 cos^3 2 θ 3 cos 2 θ =0 ⇒ cos 2 θ =0 or ±√(3)/2(i) sin 2 θ = cos 4 θ ⇒ 2 sin^2 2 θ + sin 2θ 1=0 ⇒ (2 sin 2 θ 1)(sin 2 θ+1)=0 ⇒ ...

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Byju's Answer

Standard XI

Mathematics

Principal Solution of Trigonometric Equation

The number of...

Question

The number of values of $θ$ in the interval $(- \frac{π}{2}, \frac{π}{2})$ such that $θ \neq \frac{n π}{5} f o r n = 0, \pm 1, \pm 2$ and $t a n θ = c o t 5 θ$ as well as $s i n 2 θ = c o s 4 θ$ is___

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Solution

$t a n θ = c o t 5 θ \Rightarrow c o s 6 θ = 0 \Rightarrow 4 c o s^{3} 2 θ - 3 c o s 2 θ = 0 \Rightarrow c o s 2 θ = 0 o r \pm \frac{\sqrt{3}}{2} (i) s i n 2 θ = c o s 4 θ$
$\Rightarrow 2 s i n^{2} 2 θ + s i n 2 θ - 1 = 0 \Rightarrow (2 s i n 2 θ - 1) (s i n 2 θ + 1) = 0 \Rightarrow s i n 2 θ = - 1 o r s i n 2 θ = \frac{1}{2} \Rightarrow c o s 2 θ = 0 a n d s i n 2 θ = - 1 \Rightarrow 2 θ = - \frac{π}{2} \Rightarrow θ = - \frac{π}{4}$
or $c o s 2 θ = \pm \frac{\sqrt{3}}{2}, s i n 2 θ = \frac{1}{2}$
$\Rightarrow 2 θ = \frac{π}{6}, \frac{5 π}{6} \Rightarrow θ = \frac{π}{12}, \frac{5 π}{12}$
$∴ θ = - \frac{π}{4}, \frac{π}{12}, \frac{5 π}{12}$

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Similar questions

Q. The number of values of

θ

in the interval

(- \frac{π}{2}, \frac{π}{2})

such that

tan θ = cot 5 θ

as well as

cos 2 θ = sin 4 θ

Q. The number of solution(s) of

cos 2 θ = sin 4 θ

and

tan θ = cot 5 θ

for

θ \in (- \frac{π}{2}, \frac{π}{2})

Some trigonometric ratios and the interval in which $θ$ lies is given. Match the intervals with the ratios which are positive in those intervals.

$θ$ gives positive values

p. (0, $\frac{π}{2}$ ) 1. Only sin $θ$ , cosec $θ$

q. ( $\frac{π}{2}$ , $π$ ) 2. Only cos $θ$ , sec $θ$

r. $(π, \frac{3 π}{2})$ 3. Only tan $θ$ , cot $θ$

s. $(\frac{3 π}{2}, 2 π)$ 4. All sin $θ$ , cos $θ$ , tan $θ$ , cot $θ$ , sec $θ$ , cosec $θ$

Q. Find all values of

θ

in the interval

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(1 - t a n θ) (1 + t a n θ) s e c^{2} θ + 2^{t a n^{2} θ} = 0

Q. A value of

θ

in the interval

(- \frac{π}{2}, \frac{π}{2})

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(1 - t a n θ) (1 + t a n θ) s e c^{2} θ + 2 t a n^{2} θ + 2 = 0

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The number of values of θ in the interval (−π2,π2) such that θ≠nπ5forn=0,±1,±2 and tanθ=cot5θ as well as sin2θ=cos4θ is___

tanθ=cot5θ⇒cos6θ=0⇒4cos32θ−3cos2θ=0⇒cos2θ=0or±√32(i)sin2θ=cos4θ ⇒2sin22θ+sin2θ−1=0⇒(2sin2θ−1)(sin2θ+1)=0⇒sin2θ=−1orsin2θ=12⇒cos2θ=0andsin2θ=−1⇒2θ=−π2⇒θ=−π4 or cos2θ=±√32,sin2θ=12 ⇒2θ=π6,5π6⇒θ=π12,5π12 ∴θ=−π4,π12,5π12

The number of values of $θ$ in the interval $(- \frac{π}{2}, \frac{π}{2})$ such that $θ \neq \frac{n π}{5} f o r n = 0, \pm 1, \pm 2$ and $t a n θ = c o t 5 θ$ as well as $s i n 2 θ = c o s 4 θ$ is___