Trigonometry?

Question

Given that x is sufficiently small for x^3 and higher powers of x to be neglected and higher powers of x to be neglected, and that
cos x - 4 sin x = 6x,
show that a quadratic equation for x is
x^2 + 20x - 2 = 0.
Hence find an approximate value for x, giving your answer correct to 3 significant figures.

I see. That explains why i couldn't solve it. I have yet to learn maclaurin's series.

Answer 1

series question
maclaurin series for cos x = 1 - x^2/2! + ...
maclaurin series for sin x = x + ... (no need for cubic term and higher since they can be neglected)
substitue in your equation
1 - x^2/2!-4x=6x
x^2/2 +10x-1=0
x^2 + 20x-2=0
now use the quadratic formula to find x
x =( -20 + sqrt (408))/2 = 0.0995
or x =( -20 - sqrt (408))/2, which is rejected since it is too large to be neglected

Answer 2

Using the Taylor expansions of cos x and sin x and neglecting cubic or higher terms, we get
(1 - x^2/2) - 4(x) = 6x
<=> x^2/2 + 10x - 1 = 0
<=> x^2 + 20x - 2 = 0
<=> x = -10 ± √102
<=> x = -10 ± 10.0995 (4 d.p.)
Now for x to be small enough to ignore x^3 terms, we need to take the positive root, so we get
x = 0.0995 (4 d.p.)

Answer 3

solving the quadratic equation by using

(-b+sqrt(b^2-4*a*c))/2*a

here we have a=1,b=10,c=(-2)

we have

x as 0.196

also the value 0.196 satisfies the trignometric equation


hence the answer is 0.196