How do you solve #5sin(2x)=1#?

2 Answers
maganbhai P.
Mar 9, 2018

#x=kpi+alpha,kinZandalpha=tan^(-1)(5+-2sqrt6)#
OR
#x=(kpi)/2+(-1)^k*beta/2,kinZandbeta=sin^(-1)(1/5)#

Explanation:

#5sin2x=1=>sin2x=1/5=>(2tanx)/(1+tan^2x)=1/5#
#=>10tanx=1+tan^2x=>tan^2x-10tanx+1=0#
#=>tan^2x-10tanx+25-24=0#
#=>(tanx-5)^2=24=(2sqrt(6))^2#
#=>tanx-5=+-2sqrt6=>tanx=5+-2sqrt6inR#
#=>tanx=tan(tan^(-1)(5+-2sqrt6))#, let ,#alpha=tan^(-1)(5+-2sqrt6)#
#:.x=kpi+alpha,kinZandalpha=tan^(-1)(5+-2sqrt6)#
OR
#5sin2x=1=>sin2x=1/5in[-1,1]#
#=>sin2x=sin(sin^(-1)(1/5))#, let ,#beta=sin^(-1)(1/5)#
#:.2x=kpi+(-1)^k*beta,kinZand beta=sin^(-1)(1/5)#
#=>x=(kpi)/2+(-1)^k*beta/2,kinZandbeta=sin^(-1)(1/5)#

Nghi N
Mar 10, 2018

#x = 5^@77 + k180^@#
#x = 84^@23 + k180^@#

Explanation:

#sin 2x = 1/5#
Calculator and unit circle give 2 solutions:
a. #2x = 11^@54 + k360^@# -->
#x = 5^@77 + k180^@#
b. #2x = 180 - 11.54 = 168^@46 + k360^@# -->
#x = 84.23 + k180^@#

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
4330 views around the world
You can reuse this answer
Creative Commons License