Solve the following equation sin - - sin 2-

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiple of an Angle

Solve the fol...

Question

Solve the following equation
$sin α = sin 2 α .$

Open in App

Solution

$sin α - 2 cos α sin α = 0$
$sin α (1 - 2 cos α) = 0$
$sin α = 0$ or $2 cos (α) - 1 = 0$
$α = 2 π n$ and or $cos α = \frac{1}{2}$
$α = π + 2 π n$ or $α = \frac{π}{3} + 2 π n$ , and $α = \frac{4 π}{3} + 2 π n$
Solution: $x = 2 π n, x = π + 2 π n, x = \frac{π}{3} + 2 π n, x = \frac{4 π}{3} + 2 π n$ .

Suggest Corrections

Similar questions

$s i n y = x s i n (α + y), t h e n \frac{d y}{d x} i s$

cos α + cos β + cos α = 0 = sin α + sin β + sin α

prove that

{cos}^{2} α + {cos}^{2} β + {cos}^{2} α = \frac{3}{α} = {sin}^{2} α + {sin}^{2} β + {sin}^{2} α

Q. If

\frac{s i n α}{a} = \frac{c o s α}{b}, t h e n a s i n 2 α + b c o s 2 α

If $t a n α, t a n β$ are the roots of the equation $x^{2}$ + ax + b = 0 then the value of $s i n^{2} (α + β) + a s i n (α + β) c o s (α + β) + b c o s^{2} (α + β) =$

If sin $α$ +sin $β$ +sin $γ$ =0= cos $α$ +cos $β$ +cos $γ$ ,

value of $s i n^{2}$ $α$ + $s i n^{2}$ $β$ + $s i n^{2}$ $γ$

Join BYJU'S Learning Program

Solve the following equationsinα=sin2α.

sinα−2cosαsinα=0sinα(1−2cosα)=0sinα=0 or 2cos(α)−1=0α=2πn and or cosα=12α=π+2πn or α=π3+2πn, and α=4π3+2πnSolution: x=2πn,x=π+2πn,x=π3+2πn,x=4π3+2πn.

Solve the following equation
$sin α = sin 2 α .$