If tan2 -1-3 tan -3-0- then the general value of - is

The correct option is A nπ+π/4,nπ+π/3 where nϵ I tan^2 θ tan θ √(3)tan θ+√(3)=0 ⇒tan θ[tan θ 1] √(3)[tan θ 1]=0 ⇒ (tanθ 1)(tanθ √(3)) ...

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Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

If tan2 θ-1...

Question

If ${tan}^{2} θ - (1 + \sqrt{3}) tan θ + \sqrt{3} = 0$ , then the general value of $θ$ is

n π + \frac{π}{4}, n π + \frac{π}{3}

where

n ϵ I

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n π - \frac{π}{4}, n π + \frac{π}{3}

where

n ϵ I

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n π + \frac{π}{4}, n π - \frac{π}{3}

where

n ϵ I

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n π - \frac{π}{4}, n π - \frac{π}{3}

where

n ϵ I

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Solution

The correct option is A $n π + \frac{π}{4}, n π + \frac{π}{3}$ where $n ϵ I$
${tan}^{2} θ - tan θ - \sqrt{3} tan θ + \sqrt{3} = 0$
$\Rightarrow tan θ [t a n θ - 1] - \sqrt{3} [t a n θ - 1] = 0$
$\Rightarrow (tan θ - 1) (tan θ - \sqrt{3}) = 0$
$\Rightarrow tan θ = 1, tan θ = \sqrt{3}$
So, the general solution is given by
$θ = m π + \frac{π}{4}, m \in I, θ = n π + \frac{π}{3}, n \in I$

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Similar questions

Solve the equation: $\sqrt{(\frac{1}{16} + c o s^{4} x - \frac{1}{2} c o s^{2} x)} + \sqrt{(\frac{9}{16} + c o s^{4} x - \frac{3}{2} c o s^{2} x)} = \frac{1}{2}$

Q. If

| 2 sin θ - cosec θ | \geq 1

, then complete set of values of

θ

is
(where

n \in Z

)

List-I	List-II
$1) {tan}^{2} θ = 1$	a $) n π \pm π / 6$
$2) {cos}^{2} θ = \frac{1}{4}$	b $) n π \pm π / 4$
$3) {sin}^{2} θ = \frac{1}{4}$	c $) n π \pm π / 3$
4) cosec $θ = 1$	d) $n π \pm π / 2$ $e) n π \pm π / 8$

Q. Assertion :Statement 1: General solution of

\frac{tan 3 x - tan 1 x}{1 + tan 3 x tan 1 x} = 1

x = \frac{n π}{2} + \frac{π}{8}, n ϵ I

Reason: Statement 2: General solution of

tan α = 1

α = n π + \frac{π}{4}, n ϵ I .

Q. The general solution of

(\sqrt{3} - 1) sin θ + (\sqrt{3} + 1) cos θ = 2

is
(where

n \in Z

)

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If tan2θ−(1+√3)tanθ+√3=0, then the general value of θ is

The correct option is A nπ+π4,nπ+π3 where nϵItan2θ−tanθ−√3tanθ+√3=0⇒tanθ[tanθ−1]−√3[tanθ−1]=0⇒(tanθ−1)(tanθ−√3)=0⇒tanθ=1,tanθ=√3So, the general solution is given byθ=mπ+π4,m∈I,θ=nπ+π3,n∈I

If ${tan}^{2} θ - (1 + \sqrt{3}) tan θ + \sqrt{3} = 0$ , then the general value of $θ$ is