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Continue Shopping Important forms of Trigonometric Equations There are various types of trigonometric equations and there is no standard way of dealing with all such types of equations. However, there are certain methods of tackling some set varieties. We list below certain important types along with the methods to be adopted in their solutions:

1. Equations of the form a sin2x + b sin x cos x + c cos2x + d = 0

If a + d ≠ 0, then cos x = 0 does not satisfy the equation. We divide the given equation by cos2x, then equation becomes,

a tan2 x + b tan x + c + d sec2 x = 0.

Writing sec2 x = 1 + tan2 x, we get a quadratic in tan x.

If a + d = 0, then the equation becomes

–d sin2 x + b sin x cos x + c cos2 x + d = 0

⇒ (c + d) cos2 x + b sin x cos x = 0

This can be easily solved by factorization.

2. Equations of the form a sin x + b cos x + c = 0

We divide the equation by √(a2+b2), thus the equation becomes

a/√(a2+b2 ) sin x + b/√(a2+b2 ) cos x = -c/√(a2+b2 )

⇒ cos(x–α) = -c /√(a2+b2 ),

where a/√(a2+b2) = sin α and b/√(a2+b2 ) = cos α

Clearly, the equation has no real solution if |c| > √ (a2+b2) and if

|c| < √ (a2 + b2), we can solve the equation using elementary results.

Alternatively, we can convert the given equation into a quadratic in tan x/2 by writing sin x = (2 tan (x/2))/ (1 + tan2(x/2)) and

cos x = (1 - tan2(x/2))/(1 + tan2(x/2)).

3. Equations of the form a (sin x + cos x) + b sin x cos x + c = 0

We put sin x + cos x = t ⇒ 1 ± 2 sin x cos x = t2.

So, the equation reduces to a quadratic equation.

Remark: Since here squaring is involved so the students are advised to check for extraneous roots. So once you obtain the solution, you must check the solutions which satisfy the given equation.

4. Equations containing high degree terms of sin x or cos x or both

In such a case, we try to make use of various formulae to convert the equation into linear form. The formulae that are generally used include:

sin2x = (1 - cos2x)/2,

cos2x = (1 + cos 2x)/2

sin3x = (3 sin x - sin 3x)/4

cos3x = (3cos x + cos 3x)/4 etc.

For a better idea on types of questions, refer the past year papers.

Illustration: Find the general solutions of the following equations

tan pθ + cot qθ = 0

cot θ + cosec θ = √3

5 sin2 x – 7 sin x cos x + 8 cos 2x + 4 = 0

sin10 x + cos10 x = cos42x

Solution:

(1) The given equation is tan pθ + cot qθ = 0

Hence, tan pθ = - cot qθ = tan (π/2 + qθ)

This gives pθ = nπ + π/2 – qθ, n ∈ I.

Hence, (p + q)θ = (2n + 1) π/2

This gives θ as (2n + 1) π /2(p + q).

(2) cot θ + cosec θ = √3

cos θ + 1 - √3sin θ = 0

Dividing throughout by 2, we get

1/2 cos θ - √3/2 sin θ = -1/2

cos (θ + π /3) = cos 2π /3

This gives θ + π /3 = 2nπ ± 2π /3

This yields θ = 2nπ + π /3 or θ = (2n – 1)π

But if θ = (2n – 1) π, then cot and cosec are not defined and hence this case is not possible.

So, θ = 2nπ + π /3.

(3) 5 sin2 x – 7 sin x cos x + 8 cos 2x + 4 = 0

Dividing throughout by cos2 x we get,

5 tan2 x – 7 tan x + 8 cos 2x + 16 - 4 (1 + tan2 x) = 0

tan x = 3 gives x = nπ + tan-1x

and tan x = 4 gives x = mπ + tan-14, n, m ∈ I.

(4) sin10 x + cos10 x = cos42x

Using the results, sin2x = (1 - cos2x)/2 and cos2x = (1 + cos 2x)/2

Substituting these results in the above equation we get

[(1 - cos2x)/2]5 + [(1 + cos 2x)/2]5 = cos42x

This gives 1 + 5C2 cos22x + 5C4 cos4 2x = 16 cos4 2x

11 cos4 2x - 10 cos2 2x – 1 = 0

On solving this we get cos2 2x – 1 = 0

Hence sin2 2x = 0

This gives 2x = nπ or x = nπ/2, n ∈ I

Illustration: Solve the equation (cos x – sin x)(2 tan x + sec x) + 2 = 0

Solution: Rewriting the equation, we get

(cos x – sin x) (2 tan x + ( 1/cos x) + 2 = 0

⇒ {(1-tan2 (x/2)) / (1 + tan2 (x/2)) - (2 tan (x/2)) / (1+tan2 (x/2))}.{(4 tan(x/2)) /(1-tan2 (x/2)) + (1+ tan2 (x/2)) / (1 – tan2 (x/2))} + 2 = 0

Put t = tan x/2

⇒ (1 - t2 - 2t)(4t + 1 + t2 )/(1 + t2 )(1 - t2 ) + 2 = 0 ⇒ 3t4 + 6t3 – 2t – 3 = 0

⇒ 3t4 + 8t2 – 3 + 6t3 – 2t = 0

⇒ (3t2 – 1)(t2 + 3)+ 2t (3t2 – 1) = 0

⇒ (3t2 – 1)(t2 + 2t + 3) = 0

⇒ tan2 x/2 = 1/3 = (tan π/6)2

∴ x/2 = nπ ± π/6 π x = 2nπ ± π/3, n ∈ I.

Illustration: Solve sin2θ – cosθ = ¼ for θ and write the values of θ in the interval 0 < θ < 2π.

Solution: The given equation can be written as

1 – cos2θ – cos θ = ¼

⇒ cos2θ + cos θ – ¾ = 0 ⇒ 4 cos2θ + 4cos θ – 3 = 0

⇒ cos θ = (- 4 ±√(16 + 48))/8 = 1/2,-3/2.

Since cos θ = -3/2 is not possible,

cos θ = 1/2 = cos π/3 ⇒ θ = 2nπ ± π/3.

For the given interval, n = 0 and n = 1 ⇒ θ = π/3, 5π/3.

Illustration: Solve 3 cos2θ – 2√3 sin θ cos θ – 3 sin2θ = 0.
Solution: The given equation can be written as (cos θ ≠ 0).

3 tan2θ + 2√3 tan θ – 3 = 0

⇒ tan θ = (-2√3 ± √(12 + 36))/6 = (-2√3 ± 4√3)/6 = 1/√3,-√3

Either tan θ = 1/√3 = tan π/6 ⇒ θ = nπ + π/6

or tan θ = –√3 = tan (–π/3) ⇒ θ = nπ – π/3.

Illustration: Solve cos θ + cos 3θ + cos 5θ + cos 7θ = 0.

Solution: cos θ + cos 3θ+ cos 5θ + cos 5θ + cos 7θ = 0.

⇒ 2cos 4θcos 3θ + 2cos 4θ cos θ = 0

⇒ cos 4θ (2cos 2θ cos θ) = 0.

Either cos θ = 0 ⇒ θ = (2n + 1)π/2

or cos 2θ = 0 ⇒ θ = (2n + 1)π/4 or cos 4θ = 0
⇒ θ = (2n + 1) π/8.

Illustration: Solve tan x sin 2x = 2 tan x, x  [0, 2π]
Solution: tan x sin2x = 2 tan x

tan x sin2x - 2 tan x =0

tan x (sin2x - 2) = 0.

Hence, tan x =0 or sin2x -2 =0

sin 2 x = 2

sin x =± √2

Now tan x assumes the value 0 when x = 0, π, 2π.

sin x = ± √2 is not possible since the range of sine function is [-1,1]. Hence sine

cannot assume a value greater than 1or less than -1.
x = 0, π, 2π is the only solution.

Remark:

The question seems to be easy but students often commit mistakes in these type of simple questions as well. Seeing the question, you might feel like cancelling the tan function on both sides. In such a case, we would have landed with no solution to the question.

Watch this Video for more reference

Related Resources

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