The general solution of -sin-2 x - 2 cos x - 14 - 0-is when x belongs to- -n - Z-

lt;!–td border: 1px solid #ccc;br mso data placement:same cell;– gt; Given, sin^2 x 2 cos x + 14 = 0 ⇒ 4(1 cos^2x) 8 cos x + 1 = 0 ⇒ 4cos^2 x + 8cos ...

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

Byju's Answer

Standard XII

Mathematics

General Solution of tan theta = tan alpha

The general s...

Question

The general solution of $s i n^{2} x - 2 c o s x + \frac{1}{4} = 0 i s x = (n ϵ Z)$

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

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

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

Open in App

Solution

The correct option is C
$s i n^{2} x - 2 c o s x + \frac{1}{4} = 0 \Rightarrow 4 (1 - c o s^{2} x) - 8 c o s x + 1 = 0 \Rightarrow 4 c o s^{2} x + 8 c o s x - 5 = 0 \Rightarrow 4 c o s^{2} x + 10 c o s x - 2 c o s x - 5 = 0 \Rightarrow (2 c o s x + 5) (2 c o s x - 1) = 0 \Rightarrow c o s x = \frac{1}{2} = c o s \frac{π}{3} \Rightarrow x = 2 n π \pm \frac{π}{3} .$

Suggest Corrections

Similar questions

Q. The general solution of

{tan}^{2} θ = 3

(n \in Z)

Q. The general solution of

sin 2 x + sin 4 x + sin 6 x = 0

is/are

Q. The general solution of

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

is
(where

n \in Z

)

Q. General solution of

x

for which

{cos}^{2} x = \frac{sin x tan x}{6}

is :

The general solution to $s i n^{10} x + c o s^{10} x = \frac{29}{16} c o s^{4} 2 x$ is

Join BYJU'S Learning Program

The general solution of sin2x−2 cosx+14=0 is x=(nϵZ)

The correct option is C sin2x−2cos x+14=0⇒4(1−cos2 x)−8cos x+1=0⇒4 cos2x+8 cosx−5=0⇒4cos2 x+10 cos x−2 cos x−5=0⇒(2 cos x+5)(2 cos x−1)=0⇒cos x=12=cosπ3⇒x=2nπ±π3.

The general solution of $s i n^{2} x - 2 c o s x + \frac{1}{4} = 0 i s x = (n ϵ Z)$