Question

sinx cscx -cscx+2sinx=0

Answer 1

sinx cscx = 1

therefore, 1- (1/sinx)+ 2sinx=0

so, [2(sinx)^2 + sinx -1]sinx =0. we can eliminate sinx in denominator but as sinx can be zero, and then equation will then become infinity. so basically we will write a pre-note that x is unequal to any n(pi), where n is an integer.

next in the the equation we substitute sinx with any variable 't' and solve like any quadratic, then the solutions of sinx will come

2t^2 +t -1 = 0

we get t= sinx = 1/2 or -1. so x can be pi/6 or -pi/2 in th eprinciple values, or 2n(pi) added to the prinicipal values for a general set of solutions.

Answer 2

thank you. can please help me with this problem sin 3x cos x - cos 3x sin x=.5

Answer 3

okay easy.
(sin a) (cos b) + (cos a) (sin b) = sin(a+b)

so here th eaboce equation gets down to sin(3x+x)= 0.5

sin inverse of half = 4x

so 4x = 30 degrees

so x = 7.5 degrees.

Answer 4

tanx cotx -tanx + 2cotx= 0

Answer 5

try this one now its similar to the first one that you asked!! in fact exactly the same procedure!

Answer 6

sin 3x cos x - cos 3x sin x=.5
you gave me the answer to positive cos 3x not negative

Answer 7

oops sorry. my bad. then the formula condenses to sin(A-B) as you may have gusesed

so, sin(3x-x) = sin2x = 0.5

2x= 30 degrees

so x= 15 degrees.