cos θ = 0

How to find the general solution of the equation cos θ = 0?

Prove that the general solution of cos θ = 0 is θ = (2n + 1)\(\frac{π}{2}\), n ∈ Z

Solution:

According to the figure, by definition, we have,

Cosine function is defined as the ratio of the side adjacent divided by the hypotenuse.

Let O be the centre of a unit circle. We know that in unit circle, the length of the circumference is 2π.

cos θ = 0cos θ = 0

If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), and 2π.





Therefore, from the above unit circle it is clear that 

cos θ = \(\frac{OM}{OP}\)

Now, cos θ = 0

⇒ \(\frac{OM}{OP}\) = 0

⇒ OM = 0.

So when will the cosine be equal to zero?

Clearly, if OM = 0 then the final arm OP of the angle θ coincides with OY or OY'.

Similarly, the final arm OP coincides with OY or OY' when θ = \(\frac{π}{2}\), \(\frac{3π}{2}\), \(\frac{5π}{2}\), \(\frac{7π}{2}\), ……….. , -\(\frac{π}{2}\), -\(\frac{3π}{2}\), -\(\frac{5π}{2}\), -\(\frac{7π}{2}\), ……….. i.e. when θ is  an odd  multiple  of \(\frac{π}{2}\)  i.e., when θ = (2n + 1)\(\frac{π}{2}\), where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3, …….)

Hence, θ = (2n + 1)\(\frac{π}{2}\), n ∈ Z is the general solution of the given equation cos θ = 0


1. Find the general solution of the trigonometric equation cos 3x = 0

Solution:

cos 3x = 0

⇒ 3x = (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ = 0 is (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. ]

⇒ x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solution of the trigonometric equation cos 3x = 0 is x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….


2. Find the general solution of the trigonometric equation cos \(\frac{3x}{2}\) = 0

Solution:

cos 3x = 0

⇒ 3x = (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ = 0 is (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. ]

⇒ x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solution of the trigonometric equation cos 3x = 0 is x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….





3. Find the general solutions of the equation 2 sin\(^{2}\) θ + sin\(^{2}\) 2θ = 2

Solution:

2 sin\(^{2}\) θ + sin\(^{2}\) 2θ = 2                    

⇒ sin\(^{2}\) 2θ + 2 sin\(^{2}\) θ - 2  = 0

⇒ 4 sin\(^{2}\) θ cos\(^{2}\) θ - 2 (1 - sin\(^{2}\) θ) = 0

⇒ 2 sin\(^{2}\) θ cos\(^{2}\) θ - cos\(^{2}\) θ = 0

⇒ cos\(^{2}\) θ (2 sin\(^{2}\) θ - 1) = 0

⇒ cos\(^{2}\) θ (1 - 2 sin\(^{2}\) θ) = 0

⇒ cos\(^{2}\) θ cos 2θ = 0

⇒  either cos\(^{2}\) θ = 0 or, cos 2θ = 0 

⇒ cos θ = 0 or, cos 2θ = 0 

⇒ θ = (2n + 1)\(\frac{π}{2}\)  or, 2θ = (2n + 1)\(\frac{π}{2}\) i.e., θ = (2n + 1)\(\frac{π}{2}\)

Therefore, the general solutions of the equation 2 sin\(^{2}\) θ + sin\(^{2}\) 2θ = 2 are  θ = (2n + 1)\(\frac{π}{2}\) and θ = (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, …….


4. Find the general solution of the trigonometric equation cos\(^{2}\) 3x = 0

Solution:

cos\(^{2}\) 3x = 0

cos 3x = 0

⇒ 3x = (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ = 0 is (2n + 1)\(\frac{π}{2}\), where, n = 0, ± 1, ± 2, ± 3, ……. ]

⇒ x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….

Therefore, the general solution of the trigonometric equation cos 3x\(^{2}\) = 0 is x = (2n + 1)\(\frac{π}{6}\), where, n = 0, ± 1, ± 2, ± 3, …….


5. What is the general solution of the trigonometric equation sin\(^{8}\) x + cos\(^{8}\) x =  \(\frac{17}{32}\)?

Solution:

⇒ (sin\(^{4}\) x + cos\(^{4}\) x)\(^{2}\) – 2 sin\(^{4}\) x  cos\(^{4}\) x =  \(\frac{17}{32}\)

⇒ [(sin\(^{2}\) x + cos\(^{2}\) x)\(^{2}\) - 2 sin\(^{2}\) x  cos\(^{2}\) x]\(^{2}\) -  \(\frac{(2 sinx cosx)^{4}}{8}\) = \(\frac{17}{32}\)

⇒ [1-  \(\frac{1}{2}\)sin\(^{2}\) 2x ]2  -  \(\frac{1}{8}\)sin\(^{4}\) 2x = \(\frac{17}{32}\)

⇒ 32 [1- sin\(^{2}\) 2x +  \(\frac{1}{4}\) sin\(^{4}\) 2x] - 4  sin\(^{4}\) 2x = 17 

⇒ 32 - 32 sin\(^{2}\) 2x + 8 sin\(^{4}\) 2x - 4 sin\(^{4}\) 2x – 17 = 0

⇒ 4 sin\(^{4}\) 2x  - 32 sin\(^{2}\) 2x + 15 = 0

⇒ 4 sin\(^{4}\) 2x -  2 sin\(^{2}\) 2x – 30 sin\(^{2}\) 2x + 15 = 0

⇒ 2 sin\(^{2}\) 2x (2 sin\(^{2}\) 2x - 1) – 15 (2 sin\(^{2}\) 2x - 1) = 0

⇒ (2 sin\(^{2}\) 2x - 1) (2 sin\(^{2}\) 2x - 15) = 0

Therefore,

either, 2 sin\(^{2}\) 2x - 1 = 0 ……….(1) or, 2 sin\(^{2}\) 2x - 15  = 0 …………(2)

Now, from (1) we get,

 1 - 2 sin\(^{2}\) 2x = 0

⇒  cos 4x = 0 

⇒ 4x = (2n + 1)\(\frac{π}{2}\), where, n ∈ Z   

⇒ x = (2n + 1)\(\frac{π}{8}\), where, n ∈ Z

Again, from (2) we get, 2 sin\(^{2}\) 2x = 15

⇒ sin\(^{2}\) 2x =  \(\frac{15}{2}\) which is impossible, since the numerical value of sin 2x cannot  be  greater  than 1.

Therefore, the required general solution is: x = (2n + 1)\(\frac{π}{8}\), where, n ∈ Z





● Trigonometric Equations








11 and 12 Grade Math

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