Question

How do you solve `5sin x + 3 cos x = 5`?

Answer 1

Make into a quadratic in `sin x` to find solutions:

`{ (x = pi/2 + 2kpi), (x = sin^(-1)(8/17) + 2kpi) :}`

for all `k in ZZ`

Explanation:

Subtract `5 sin x` from both sides to get:

`3 cos x = 5 - 5 sin x`

Square both sides (noting that this may introduce spurious solutions) to get:

`9 cos^2 x = 25 - 50 sin x + 25 sin^2 x`

Now `sin^2 x + cos^2 x = 1`, so `cos^2 x = 1 - sin^2 x` and we get:

`9 (1 - sin^2 x) = 25 - 50 sin x + 25 sin^2 x`

Subtracting the left hand side from the right, this becomes:

`34 sin^2x - 50 sin x + 16 = 0`

Divide through by `2` to get:

`0 = 17 sin^2 x - 25 sin^2 x + 8 = (sin x - 1)(17 sin x - 8)`

So `sin x = 1` or `sin x = 8/17`

If `sin x = 1` then `color(blue)(x = pi/2 + 2kpi)` for any `k in ZZ`

These are valid solutions since `cos (pi/2+2kpi) = 0`

How about `x = sin^(-1) (8/17)` ?

`5 - 5 sin x = 5 - 40/17 = (85-40)/17 = 45/17`

So `(5 - 5 sin x)/3 = 15/17`

With `sin x = 8/17` and `cos x = 15/17` we require a value of `x` in Q1 plus any integer multiple of `2pi`.

So we have solutions: `color(blue)(x = sin^(-1)(8/17) + 2kpi)` for any `k in ZZ`