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 = 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
11 and 12 Grade Math
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