GENERAL SOLUTION OF SOME SIMPLE EQUATION

sin Θ = 0                                   →  Θ = nΠ

cos Θ = 0                                  → Θ = (2n + 1) Π/2

tan Θ = 0                                  → Θ = nΠ

cot Θ = 0                                 → Θ = (2n + 1)Π/2

sin Θ = 1                                  → Θ = (4n + 1) Π/2

sin Θ = -1                                → Θ = (4n + 3) Π/2

cos Θ = 1                                → Θ = 2n Π

cos Θ = -1                              → Θ = (2n + 1) Π

tan Θ = not defined           → Θ = (2n + 1) Π/2

cot Θ = Not defined          → Θ = nΠ

cosec Θ = Not defined     → Θ = nΠ

sec Θ = not defined            → Θ= (2n + 1) Π/2

ü      There are some cautions to be taken while solving some Trigonometric Equations. They are listed down here.

Þ   Check the validity of the given equation

e.g., 2sin Θ – cos Θ = 4 can never be true for any Θ as the value (2 sin Θ – cos Θ) can never exceed √(2²+(-1)² = √5. So there is no solution to this equation.

Þ   Equation involving sec Θ and / or tan Θ can never have a solution of the form Θ = (2n + 1) Π/2. similarly, equation involving cosec Θ and / or cot Θ can never have solution of the form Θ = nΠ. The corresponding function are undefined at these values of Θ.

Þ   Avoid squaring the equations as far as possible because it leads to extraneous solution. If it has to be squared, check whether the solution (s) arrived at satisfy the original unsquared equation or not. e.g., given that x = 4 → x² = 16 → x = ± 4. But originally x = 4 only.

Þ   Do not cancel common factor involving the unknown angle on L.H.S. and R.H.S because it may delete some solution. e.g. In the equation sin Θ (2cos Θ – 1) = sin Θ cos²Θ if we cancel sin Θ on both sides we get cos² Θ – 2cos Θ + 1 = 0 → (cos Θ – 1)² = 0 → cos Θ = 1 → Θ = 2n Π.

Þ   But Θ = nΠ also satisfies the equation because it makes sin Θ = 0. So, the complete solution is Θ = nΠ, n Î Z.

Þ   Denominator terms of the equation if present should never become zero at any stage while solving for any value of Θ contained in the answer.

Þ        Something the equation has some limitation also. e.g., cot² Θ cosec² Θ = 1 can be true only if cot² Θ = 0 and cosec² Θ = 1 simultaneously as                 cosec² Θ ≥ 1. Hence the solution is Θ= (2n + 1) Π/2.

Illustrations

1. Obtain the general solution of secx + tanx =√3

Sol. Since secx and tanx. Both are undefined for x = (2n + 1)Π/2, the final solution should not include any odd multiple of Π/2.

Now,√3 cosx – sin x = 1 => cos(Θ + Π/6) = ½ = cosΠ/3

Þ Θ-+ Π/6 = 2nΠ + Π/3.

With ‘+’ sign, it given Θ = 2nΠ + Π/6

With ‘-‘ sign, it given Θ = 2nΠ – Π/2 = (4n – 1)Π/2

Since (4n -1) is always odd therefore we should delete this solution.

Thus, Θ= (12n + 1)Π/6, n Î Z.