# 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.

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 = 1

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

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