How do you solve 2sinx-3cosx=1?

Answer 1

2sin(x) - 3cos(x) = 1
2sin(x) - 1 = 3cos(x)
2sin(x) - 1 = 3 √[ 1 - sin^2 (x)]
4sin^2(x) - 4sin(x) + 1 = 9(1 - sin^2(x))
13sin^2(x) - 4sin(x) - 8 = 0
sin(x) = [ 4 ± √(16 - 4(13)(-8)) ] / 26
sin(x) = [ 4 ± 12√3] / 26
sin(x) = [ 2 ± 6√3] / 13
x = sin^-1 ([2+6√3] / 13), sin^-1 ([2-6√3] / 13)

Answer 2

3 sin(2x) = 2 sinx 3 (2 sinx cosx) = 2 sinx 6 sinx cosx = 2 sinx Do NOT divide both sides by sinx. This will eliminate a valid answer. Instead factor: 6 sinx cosx - 2 sinx = 0 2 sinx = (3 cosx - 1) = 0 First we'll solve on interval [0°, 360°] sinx = 0 ------> x = 0°, 180°, 360° cosx = 1/3 ---> x = 70.53°, 289.47° On interval (-∞, ∞) x = 180°k x = 70.53° + 360°k x = 289.47 + 360°k, for all integers k Mαthmφm

Answer 3

Write the LHS as
2sinx-3cosx = R*sin(x - θ)
= R*cosθ*sinx - R*sinθ*cosx

Clearly R*cosθ = 2
R*sinθ = 3
tanθ = 3/2
θ = 56.3 deg
R^2 = 2^2 + 3^2 = 13

So now we have
sqrt(13) * sin(x - θ) = 1
sin(x - θ) = 1/sqrt(13)
x - θ = 16.1, 163.9 deg
x = 72.41, 220.21 deg

**EDIT**
There is just one small problem with Geezah's approach. It's not wrong, but when you square both sides like that, you introduce solutions which do not fit the original problem.
His answer is
x = sin^-1 ([2+6√3] / 13), sin^-1 ([2-6√3] / 13)
= asin(0.9533), asin(-0.6456)
= 72.41, -40.21 or 319.79
Other solutions between zero and 360 are
107.59, 220.21

If you take x = 107.59 and 319.79 (or -40.21), they don't fit the original equation.
2*sin(107.59) - 3*cos(107.59) = 2.813
2*sin(319.79) - 3*cos(319.79) = -3.582
However if you change the equation to 2sinx + 3cosx, you'll get 1 on the RHS. That's the danger of squaring both sides.

Answer 4

2sinx-3cosx
= √(13)[sinx 2/√(13) - cosx 3/√(13)]
= √(13)sin[x-arctan(3/2)]
= 1
x = 72.4120, 220.2078 degrees

Answer 5

Offhand, I dont remember any handy-dandy substitutions, so you can always do a trial an error approach. One solution seems to occur between pi/3 and pi/2