HOW DO YOU SOLVE 2COS(3X + PI/4) =1?

Here we will use the following formulas to lớn get the solution of the trigonometric equations.

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(a) If sin θ = 0 then θ = nπ, where n = 0, ± 1, ± 2, ± 3, …….

(b) If cos θ = 0 then θ = (2n + 1) (fracπ2), where n = 0, ± 1, ± 2, ± 3, …….

(c) If cos θ = cos ∝ then θ = 2nπ ± ∝, where n = 0, ± 1, ± 2, ± 3, …….

(d) If sin θ = sin ∝ then θ = n π + (-1) (^n) ∝, where n = 0, ± 1, ± 2, ± 3, …….

(e) If a cos θ + b sin θ = c then θ  = 2nπ + ∝ ±  β, where cos β = (fraccsqrta^2  +  b^2), cos ∝ = (fracasqrta^2  +  b^2) và sin ∝ = (fracbsqrta^2  +  b^2), where n = 0, ± 1, ± 2, ± 3, …….


1. Solve tung x + sec x = √3. Also find values of x between 0° và 360°.

Solution:

tan x + sec x = √3

⇒ (fracsin xcos x) + (frac1cos x) = √3, where cos x ≠ 0

⇒ sin x + 1 = √3 cos x

⇒ √3 cos x - sin x = 1,

This trigonometric equation is of the size a cos θ + b sin θ = c where a = √3, b = -1 và c = 1.

⇒ Now dividing both sides by (sqrt(sqrt3)^2 + (1)^2)

⇒ (frac√32) cos x - (frac12)sin x = (frac12)

⇒ cos x cos (fracπ4) – sin x sin (fracπ6) = cos (fracπ3)

⇒ cos (x + (fracπ6)) = cos (fracπ3)

⇒ x + (fracπ6) = 2nπ ± (fracπ3), where n = 0, ± 1, ± 2, ± 3, …….

⇒ x = 2nπ ± (fracπ3) - (fracπ6), where n = 0, ± 1, ± 2, ± 3, …….

When we take minus sign with (fracπ3), we get

x = 2nπ - (fracπ3) - (fracπ6)

⇒ x = 2nπ - (fracπ2), so that cos x = cos (2nπ - (fracπ2)) = cos (fracπ2) = 0, which spoils the assumption cos x  ≠ 0 (otherwise the given equation would be meaningless).

So, x = 2nπ + (fracπ3) - (fracπ6), where n = 0, ± 1, ± 2, ± 3, …….

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⇒ x = 2nπ + (fracπ6), where, n = 0, ± 1, ± 2, ± 3, ……. Is the general

solution of the given equation tan x + sec x = √3.

The only solution between 0° and 360° is x = (fracπ6) = 30°

2. Find the general solutions of θ which satisfy the equation sec θ = - √2

Solution:   

sec θ = -  √2

⇒ cos θ = - (frac1√2)

⇒ cos θ = - cos (fracπ4)

⇒ cos θ = cos (π - (fracπ4))

⇒ cos θ = cos (frac3π4)

⇒ θ = 2nπ ± (frac3π4), where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solutions of θ which satisfy the equation sec θ = - √2 is θ = 2nπ ± (frac3π4), where, n = 0, ± 1, ± 2, ± 3, …….

3. Solve the equation 2 cos(^2) x + 3 sin x = 0

Solution:

2 cos(^2) x + 3 sin x = 0

⇒ 2(1 - sin(^2) x) + 3 sin x = 0

⇒ 2 – 2 sin(^2) x + 3 sin x = 0

⇒ 2 sin(^2) x – 3 sin x – 2 = 0

⇒ 2 sin(^2) x - 4 sin x + sin x – 2 = 0

⇒ 2 sin x(sin x - 2) + 1(sin – 2) = 0

⇒ (sin x - 2)(2 sin x + 1) = 0

⇒ Either sin x - 2 =0 or 2 sin x + 1 = 0

But sin x – 2 = 0 i.e., sin x = 2, which is not possible.

Now khung 2 sin x + 1 = 0 we get

⇒ sin x = -½ 

⇒ sin x =- sin (fracπ6)

⇒ sin x = sin (π + (fracπ6))

⇒ sin x = sin (frac7π6)

⇒ x = nπ + (1)(^n)(frac7π6), where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the solution for the equation 2 cos(^2) x + 3 sin x = 0 is x = nπ + (1)(^n)(frac7π6), where, n = 0, ± 1, ± 2, ± 3, …….Note: In the above trig equation we observe that there is more than one trigonometric function. So, the identities (sin (^2) θ + cos (^2) θ = 1) are required to reduce the given equation to lớn a single function.

4. Find the general solutions of cos x + sin x = cos 2x + sin 2x

Solution:

cos x + sin x = cos 2x + sin 2x

⇒cos x - cos 2x - sin 2x + sin x = 0

⇒  (cos x - cos 2x) - (sin 2x - sin x) = 0

⇒  2 sin (frac3x2) sin (fracx2) - 2 cos (frac3x2) sin (fracx2) = 0

⇒  sin (fracx2)  (sin (frac3x2) - cos (frac3x2)) = 0 Therefore, either, sin (fracx2) = 0          

⇒ (fracx2)= nπ     

⇒ x = 2nπ

or, sin (frac3x2) -  cos (frac3x2) = 0

⇒ sin (frac3x2) = cos (frac3x2)

⇒ chảy (frac3x2) = 1

⇒ chảy (frac3x2) = tan (fracπ4)

⇒ (frac3x2)= nπ + (fracπ4)

⇒ x = (frac13) (2nπ + (fracπ2)) = (4n + 1)(fracπ6)Therefore, the general solutions of cos x + sin x = cos 2x + sin 2x are x = 2nπ & x = (4n+1)(fracπ6), Where, n = 0, ±1, ±2, ………………….. 5. Find the general solutions of sin 4x cos 2x = cos 5x sin x

Solution:

sin 4x cos 2x = cos 5x sin x

⇒ 2 sin 4x cos 2x = 2 cos 5x sin x

⇒ sin 6x + sin 2x = sin 6x - sin 4x     

⇒ sin 2x + sin 4x =0

⇒ 2sin 3x cos x =0Therefore, either, sin 3x = 0 or, cos x = 0

i.e., 3x = nπ or, x = (2n + 1)(fracπ6)

⇒ x = (fracnπ3) or,  x = (2n + 1)(fracπ6)Therefore, the general solutions of sin 4x cos 2x = cos 5x sin x are (fracnπ3) và x = (2n + 1)(fracπ6)

 Trigonometric Equations

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