How do you find all solutions of the equation $cos(x+pi/6)-cos(x-pi/6)=1$ in the interval $[0,2pi)$ ?

Question

How do you find all solutions of the equation $cos(x+pi/6)-cos(x-pi/6)=1$ in the interval $[0,2pi)$ ?

2 Answers

Impact of this question

8964 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-01T19:39:39

See below.

Explanation:

$cos(x + a) - cos(x - a) = -2 sin x sin a$

so

$-2 sin x sin a=1->sinx = -1/(2sina) = -1/(2 xx 1/2) = -1$

so

$x = arcsin(-1)+2kpi=- pi/2+2kpi$ then

$x = 3/2pi$

Answer 2 · 2017-03-02T00:34:27

$x = (3pi)/2$

Explanation:

Here's a different way of proceeding. Use the sum and difference formulae $cos(A + B) = cosAcosB - sinAsinB$ and $cos(A - B) = cosAcosB + sinAsinB$ .

$cosxcos(pi/6) - sinxsin(pi/6) - (cosxcos(pi/6) + sinxsin(pi/6)) = 1$

$cosx(sqrt(3)/2) - sinx(1/2) - (cosx(sqrt(3)/2) + sinx(1/2)) = 1$

$sqrt(3)/2cosx - 1/2sinx - sqrt(3)/2cosx - 1/2sinx = 1$

$-sinx = 1$

$sinx = -1$

$x = (3pi)/2$

Hopefully this helps!

Science

Math

Humanities

... and beyond

How do you find all solutions of the equation $cos(x+pi/6)-cos(x-pi/6)=1$ in the interval $[0,2pi)$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find all solutions of the equation cos(x+pi/6)-cos(x-pi/6)=1cos(x+pi/6)-cos(x-pi/6)=1 in the interval [0,2pi)[0,2pi)?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you find all solutions of the equation $cos(x+pi/6)-cos(x-pi/6)=1$ in the interval $[0,2pi)$ ?