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

# The general solution of the equation sin2θ=sin2α is/are

A

nπ+α, nZ.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B

nπα, nZ.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C

(2n+1)π2α, nZ.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

(2n+1)π2+α, nZ.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct options are A nπ+α, n∈Z. B nπ−α, n∈Z.Here, in order to find the solution of the equation, we will express the equation in linear powers i.e. we can write the equation as: sin2θ=sin2α(sinθ− sinα)(sinθ+sinα)=04×(sin(θ−α)2cos(θ+α)2)(sin(θ+α)2cos(θ−α)2)=0sin(θ−α)×sin(θ+α)=0sin(θ−α) =0 or sin(θ+α)=0(θ−α)= nπ or (θ+α)= nπ⇒θ=nπ+α or θ=nπ−α, n∈Z⇒θ=nπ±α , n∈Z

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Solving Trigonometric Equations
MATHEMATICS
Watch in App
Join BYJU'S Learning Program