The correct option is A θ =nπ ( 1 ) #160;^ n ( π #160;/ 6 #160;) Given, #160;2sinθ +1 =0⇒sinθ = 1 / 2 ⇒θ =nπ +( 1 ) #160;^ n ( ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

Solve 2sinθ...

Question

Solve $2 sin θ + 1 = 0$

θ = n π - {(- 1)}^{n} (\frac{π}{6})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

θ = n π + {(- 1)}^{n} (\frac{π}{6})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

θ = 2 n π \pm \frac{π}{6}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

θ = 2 n π \pm \frac{π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $θ = n π - {(- 1)}^{n} (\frac{π}{6})$
Given, $2 sin θ + 1 = 0 \Rightarrow sin θ = \frac{- 1}{2}$

$\Rightarrow θ = n π + {(- 1)}^{n} (\frac{- π}{6})$

$\Rightarrow θ = n π - {(- 1)}^{n} (\frac{π}{6})$

Suggest Corrections

Similar questions

Consider the system of linear equations in x, y and z ;
(sin 3 $θ$ ) x - y + z = 0 ......(i)
(cos 2 $θ$ )x + 4y + 3z = 0 ......(ii)
2x + 7y+ 7z = 0 ......(iii)

The value of $θ$ for which the system has nontrivial solution is

Q. The general solution of

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

is
(where

n \in Z

)

Q. The general solution of

t a n^{2} θ = 3 i s (n ϵ Z)

Q. The general solution of

tan x + tan 2 x + \sqrt{3} tan x \cdot tan 2 x = \sqrt{3}

(where n \in Z)

Q. The general values of

θ

satisfying the equation

2 s i n^{2} θ - 3 s i n θ - 2 = 0

(n \in Z)

Join BYJU'S Learning Program

Solve 2sinθ+1=0

The correct option is A θ=nπ−(−1)n(π6)Given, 2sinθ+1=0⇒sinθ=−12⇒θ=nπ+(−1)n(−π6)⇒θ=nπ−(−1)n(π6)

Solve $2 sin θ + 1 = 0$

The correct option is A $θ = n π - {(- 1)}^{n} (\frac{π}{6})$
Given, $2 sin θ + 1 = 0 \Rightarrow sin θ = \frac{- 1}{2}$

$\Rightarrow θ = n π + {(- 1)}^{n} (\frac{- π}{6})$

$\Rightarrow θ = n π - {(- 1)}^{n} (\frac{π}{6})$