for trigometry..?

Question

if given trigo questionss example : 3 cos x + sin x =1 ..how would we knoe which way to use to solve this?..using normal way?..or using substitution t= tan x/2 or using a cos x + b sin x = c form?

Answer 1

Let
3cos x + sin x = k cos (x - a) where k > 0
3cos x + sin x = (k cosa) cos x + (k sin a) sin x
3 = k cos a
1 = k sin a
3² + 1² = k²(cos² a + sin² a)
10 = k²
k = √10

sin a / cos a = tan a = 1 / 3
a = 18.4°
√10 cos (x - 18.4)° = 1
cos (x - 18.4)° = 1 / √10
x - 18.4 = 71.6° or 288.4°
x = 90° or 307° ( to nearest whole number)

Answer 2

3cos(x) + sin(x) = 1
Square both sides:
9cos^2(x) + 6cos(x)sin(x) + sin^2(x) = 1
8cos^2(x) + cos^2(x) + 6cos(x)sin(x) + sin^2(x) = 1
8cos^2(x) + 6cos(x)sin(x) + 1 = 1
8cos^2(x) + 6cos(x)sin(x) = 0
2cos(x) [4cos(x) + 3sin(x)] = 0

2cos(x) = 0
cos(x) = 0
x = kπ/2, where k is an odd integer

4cos(x) = -3sin(x)
-4/3 = tan(x)
x = arctan(-4/3)

If you check both solutions, they both satisfy the equation.

So x = kπ/2, where k is an odd integer and arctan(-4/3)

Answer 3

think of a top triangle who hypotenuse is c, legs a and b. the sin is a/c, the cos is b/c. Your difficulty then is (a/c)/(b/c). Invert the denominator and multiply a/c * c/b you get ac/bc. The c's cancel and you're left with a/b that's the tangent. i'm hoping it is clean. Trig is all approximately ratios - get those firmly in techniques and each thing falls into place. good success!

Answer 4

Use whatever one will work and you can prove.