# precalulas

posted by .

solve each equation over the given interval 2x=cosx for all values of x

• precalulas -

There is no easy method to solve equations of this type.
One way is to graph the left side and the right side as individual functions and see where they intersect
e.g.
graph : y = 2x and y = cosx
you will see that they intersect appr. at x = 1/2

As a matter of fact that is how "Wolfram" shows the solution
http://www.wolframalpha.com/input/?i=2x%3Dcos%28x%29+

With a good calculator at hand, you can then pinpoint the answer more accurately
e.g.
x = .5, LS = 1, RS = .877 , RS < LS so lower the x
x = .4, LS = .8, RS = .921 RS > LS so raise the x
x = .45 , LS = .9, RS = .90044 , not bad , RS > LS , so raise the x a bit
x = .452, LS = .904 , RS = .8995 , RS < LS, lower the x
x = .4505, LS = .901 , RS = .9002
do you get the idea.

Wolfram had it at x = .450184
then LS = .900368
RS = .900367 , how is that for close

## Similar Questions

1. ### Maths

Solve this equation fo rx in the interval 0<=x<=360 3sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3 sin^2x/cosx = 8/3 (1-cos^2x)/cosx=8/3 cross-multiply 3 - 3cos^2x = 8cosx 3cos^2x + 8cosx - 3 = 0 (3cosx-1)(cosx+3)=0 …
2. ### math/trigonometry

solve secx/2=cosx/2 over the interval[0,2pi) without using a calculator
3. ### CACULUS

THE VALUES OF X THAT ARE SOLUTIONS TO THE EQUATION [COSX]^2 = SIN2X IN THE INTERVAL 0 TO PI?
4. ### maths

q1) 2cos2x is f(x) -square root3 is g(x) solve f(x)-g(x)=0 to get points of intersection in x is between 0 and 180degrees q2) a) write the equation cos2x + 8cosx+9=0 in terms of cosx and show that for cosx it has equal roots q2b) show …
5. ### Calculus

solve this equation on the given interval expressing the solution for x in terms of inverse trigonometric functions. 6(cosx)^2-cosx-5=0 on (pie/2, pie)
6. ### Precalculus

Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π). 8. cos2x=cosx 10. 2cos^2x+cosx=cos2x Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your …
7. ### Trigonometry

Solve the equation for all values of x. -2cosĀ²x-sin x+1=0, on the interval [0,2π).
8. ### Trigonometry

Solve the equation for all values of x. 2sin(2x)-√3=0, on the interval [0,2π).
9. ### math

verify that the function satisfies the hypotheses of the mean values theorem on the given interval. then find all numbers c in the given interval that satisfy the conclusion of the mean values theorem f(x)= 2x/x-4 [0,3]
10. ### Math - extreme values

Identify the critical points and find the extreme values on the interval [-1,-5) for f(x)=cosx+xsinx+3 I've taken the derivative which gives me f'(x)=xcos(x). I know I have to solve to get the critical points and then plug in the critical …

More Similar Questions