177k views
0 votes
How do I solve?
5 sin 4x = −10 sin 2x

2 Answers

5 votes
Answer:

Solve 5sin 4x = -10sin 2x

Ans: 0, pi/2, pi

Step-by-step explanation:

sin 4x = 2sin 2x.cos 2x
10sin 2x.cos 2x + 10sin 2x = 0
10sin 2x(cos 2x + 1) = 0
a. sin 2x = 0 -->
2x = 0 --> x = 0
2x = pi --> x = pi/2
2x = 2pi x = pi
b. cos 2x = -1 --> 2x = pi --> x = pi/2

User Rocambille
by
7.1k points
4 votes
Move 5cos(x)5cos(x) to the left side of the equation by subtracting it from both sides.5sin(2x)−5cos(x)=05sin(2x)-5cos(x)=0Simplify each term.Apply the sine double-angle identity.5(2sin(x)cos(x))−5cos(x)=05(2sin(x)cos(x))-5cos(x)=0Multiply 22 by 55 to get 1010.10(sin(x)cos(x))−5cos(x)=010(sin(x)cos(x))-5cos(x)=0Remove parentheses around sin(x)cos(x)sin(x)cos(x).10sin(x)cos(x)−5cos(x)=010sin(x)cos(x)-5cos(x)=0Factor 5cos(x)5cos(x) out of 10sin(x)cos(x)−5cos(x)10sin(x)cos(x)-5cos(x).
Factor 5cos(x)5cos(x) out of 10sin(x)cos(x)10sin(x)cos(x).5cos(x)(2sin(x))−5cos(x)=05cos(x)(2sin(x))-5cos(x)=0Factor 5cos(x)5cos(x) out of −5cos(x)-5cos(x).5cos(x)(2sin(x))+5cos(x)(−1)=05cos(x)(2sin(x))+5cos(x)(-1)=0Factor 5cos(x)5cos(x) out of 5cos(x)(2sin(x))+5cos(x)⋅−15cos(x)(2sin(x))+5cos(x)⋅-1.5cos(x)(2sin(x)−1)=0
User Lejonl
by
6.7k points