Having a problem with answering a trigonometric function

Question

I am stuck at this question I think it's easy but I am just looking at it the wrong way:

This question is on angle sum formula

3sin(x)+4cos(x) = 5 then 3cos(x) - 4sin(x) = .......

Answer 1

Let a = sin(x) b = cos(x) & c = unknown answer Then

3a + 4b = 5 Multiply equation 1 by 4 & equation 2 by 3 and get 12a + 16b = 20

-4a + 3b = c -12 + 9b = 3c

adding the equations gives 25b = 20 + 3c or c = (25b -20)/3 or

c = (25cos(x) -20)/3

Answer 2

3*sin(x) + 4*cos(x) = 5

3 * (1 - cos^2(x)) + 4 * cos(x) = 5

3 - 3cos^2 + 4cos - 5 = 0

-3 cos^2 + 4 cos - 2 = 0

3cos^2 - 4 cos + 2 = 0

M^2 - 4M + 2 = 0 where M = cos

M = [ 4 +or- sqrt( 16 - 8)]/ 2

= [ 4 +or- sqrt(8)]/2

= [ 4 +or- 2 * sqrt(2)] / 2

= 2 +or- sqrt(2)

so x = arccos( 2 + sqrt(2) ) or x = arccos( 2 - sqrt(2))

after calculating those, you can find 3cos(x) - 4sin(x)

Note that the first one will not work because cos and sin are always between

1 and -1, and 2 +sqrt(2) > 1

so x = arccos(2 - sqrt(2)) = 54.141432378941....