Find all the values of x in between the range [0,2π] for the following equation 2sin(4x)-3cos(4x)=0
Answers
Answer:
sin
2
x+cos
2
x=1
sin
4
x+cos
4
x+2sin
2
xcos
2
x=1
sin
4
x+cos
4
x=1−2sin
2
xcos
2
x ...[1]
sin
2
x+cos
2
x=1
sin
6
x+cos
6
x+3sin
2
xcos
2
x(sin
2
x+cos
2
x)=1
sin
6
x+cos
6
x=1−3sin
2
xcos
2
x ...[2]
∣
2sin
4
x+18cos
2
x
−
2cos
4
x+18sin
2
x
∣=1
Squaring above equation
∴2sin
4
x+18cos
2
x+2cos
4
x+18sin
2
x−2(
2sin
4
x+18cos
2
x
)(
2cos
4
x+18sin
2
x
)=1
∴2(sin
4
x+cos
4
x)+18(cos
2
x+sin
2
x)−1=2(
2sin
4
x+18cos
2
x
)(
2cos
4
x+18sin
2
x
)
∴2(1−2sin
2
xcos
2
x)+18−1=2(
2sin
4
x+18cos
2
x
)(
2cos
4
x+18sin
2
x
) ..( [Using[1] )
∴19−4sin
2
xcos
2
x=2(
2sin
4
x+18cos
2
x
)(
2cos
4
x+18sin
2
x
)
Squaring above equation
∴361+16sin
4
xcos
4
x−152sin
2
xcos
2
x=4(4sin
4
xcos
4
x+36cos
6
x+36sin
6
x+324sin
2
xcos
2
x)
∴361+16sin
4
xcos
4
x−152sin
2
xcos
2
x=16sin
4
xcos
4
x+144cos
6
x+144sin
6
x+1296sin
2
xcos
2
x
∴361=144(cos
6
x+sin
6
x)+1448sin
2
xcos
2
x
∴361=144(1−3sin
2
xcos
2
x)+1448sin
2
xcos
2
x ....( Using [2] )
∴217=1016sin
2
xcos
2
x
∴217=254sin
2
2x
∴sin2x=±
254
217
Let sin
−1
254
217
=θ
sin2x=±
254
217
∴2x=sin
−1
254
217
∴2x=θ,π−θ,2π+θ,3π−theta
∴x=
2
θ
,
2
π−θ
,
2
2π+θ
,
2
3π−θ
2x=sin
−1
(−
254
217
)
∴2x=π+θ,2π−θ,3π+θ,4π−θ
∴x=
2
π+θ
,
2
2π−θ
,
2
3π+θ
,
2
4π−θ
So, total number of solutions =8
So, the answer is option (D)
Step-by-step explanation: