Question

Solve 10cos x+13cos x/2=5 ?

Answer 1

Solution: `(x ~~106.26^0 , x ~~ -106.26^0)`

Explanation:

`10 cos x +13 cos(x/2) =5 ; [cos x = 2 cos ^2 (x/2) -1]` or

`10(2 cos ^2 (x/2) -1) +13 cos (x/2) -5=0`

`20 cos ^2 (x/2) +13 cos (x/2) -15=0` or

`20 cos ^2 (x/2) +25 cos (x/2)- 12 cos (x/2) -15=0` or

`5 cos(x/2)(4 cos (x/2)+5)-3(4 cos (x/2) +5)=0` or

`(4 cos (x/2)+5)(5 cos (x/2) -3)=0 :.` Either

`(4 cos (x/2)+5)=0 or (5 cos (x/2) -3)=0`

`(4 cos (x/2)+5)=0 :. 4 cos (x/2)=-5` or

`cos (x/2) !=5/4` since range of `cos x` is `[-1,1]`

`(5 cos (x/2) -3)=0 :. 5 cos (x/2) =3` or

`cos (x/2) =3/5:. x/2 = cos ^-1 (3/5)~~ 53.13^0`

Also `cos (-53.13)~~3/5 :. x= 53.13*2 ~~106.26^0`

and `x= (-53.13)*2 ~~ -106.26^0`

Solution: `(x ~~106.26^0 , x ~~ -106.26^0)` [Ans]

Answer 2

`x = \pm arccos(-7/25) + 4 pi k quad` integer `k`

Explanation:

Let's start by substituting `y=x/2` to get rid of the fractional angles.

`10 cos(2y) + 13 cos y = 5`

The favored form of the cosine double angle formula is

`cos(2y) = 2 cos^2 y -1`

Substituting,

`10(2 cos^2 y - 1) + 13 cos y - 5 = 0`

`20 cos^2y + 13 cos y - 15 = 0`

That's a pain to factor but a little search comes up with

`(5 cos y - 3)(4 cos y + 5) = 0`

`cos y = 3/5 or cos y = -5/4`

We can ignore the out of range cosine.

`cos y = 3/5`

We can use the double angle formula:

`cos x =cos(2y) =2 cos^2 y - 1 = 2(3/5)^2-1=-7/25`

`x = arccos(-7/25)`

That's a Pythagorean Triple `7^2+24^2=25^2` so we can try to write that as `arctan(\pm 24/7)` but that's more extraneous roots.

`x = \pm arccos(-7/25) + 2 pi k quad` integer `k`

Check:

We'll check a couple with a calculator.

`x = text{Arc}text{cos}(-7/25) approx 106.260205 ^circ`

# 10 cos(106.260205) + 13 cos(106.260205/2)-5 = -7 times 10^{-8} quad sqrt#

Let's add 360 and check again:

# 10 cos(360+106.260205) + 13 cos( (360+106.260205/2 ) )-5 = -15.6 quad# DOESN'T WORK.

Because of the half angle, the correct answer seems to be

`x = \pm arccos(-7/25) + 4 pi k quad` integer `k`