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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)π2, 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, π2, π, 3π2, and 2π.
Therefore, from the above unit circle it is clear that
cos θ = OMOP
Now, cos θ = 0
⇒ OMOP = 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 = π2, 3π2, 5π2, 7π2, ……….. , -π2, -3π2, -5π2, -7π2, ……….. i.e. when θ is an odd multiple of π2 i.e., when θ = (2n + 1)π2, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3, …….)
Hence, θ = (2n + 1)π2, 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)π2, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ = 0 is (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation cos 3x = 0 is x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
2. Find the general solution of the trigonometric equation cos 3x2 = 0
Solution:
cos 3x = 0
⇒ 3x = (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ = 0 is (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation cos 3x = 0 is x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
3. Find the general solutions of the equation 2 sin2 θ + sin2 2θ = 2
Solution:
2 sin2 θ + sin2 2θ = 2
⇒ sin2 2θ + 2 sin2 θ - 2 = 0
⇒ 4 sin2 θ cos2 θ - 2 (1 - sin2 θ) = 0
⇒ 2 sin2 θ cos2 θ - cos2 θ = 0
⇒ cos2 θ (2 sin2 θ - 1) = 0
⇒ cos2 θ (1 - 2 sin2 θ) = 0
⇒ cos2 θ cos 2θ = 0
⇒ either cos2 θ = 0 or, cos 2θ = 0
⇒ cos θ = 0 or, cos 2θ = 0
⇒ θ = (2n + 1)π2 or, 2θ = (2n + 1)π2 i.e., θ = (2n + 1)π2
Therefore, the general solutions of the equation 2 sin2 θ + sin2 2θ = 2 are θ = (2n + 1)π2 and θ = (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, …….
4. Find the general solution of the trigonometric equation cos2 3x = 0
Solution:
cos2 3x = 0
cos 3x = 0
⇒ 3x = (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation cos θ
= 0 is (2n + 1)π2, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation cos 3x2 = 0 is x = (2n + 1)π6, where, n = 0, ± 1, ± 2, ± 3, …….
5. What is the general solution of the trigonometric equation sin8 x + cos8 x = 1732?
Solution:
⇒ (sin4 x + cos4 x)2 – 2 sin4 x cos4 x = 1732
⇒ [(sin2 x + cos2 x)2 - 2 sin2 x cos2 x]2 - (2sinxcosx)48 = 1732
⇒ [1- 12sin2 2x ]2 - 18sin4 2x = 1732
⇒ 32 [1- sin2 2x + 14 sin4 2x] - 4 sin4 2x = 17
⇒ 32 - 32 sin2 2x + 8 sin4 2x - 4 sin4 2x – 17 = 0
⇒ 4 sin4 2x - 32 sin2 2x + 15 = 0
⇒ 4 sin4 2x - 2 sin2 2x – 30 sin2 2x + 15 = 0
⇒ 2 sin2 2x (2 sin2 2x - 1) – 15 (2 sin2 2x - 1) = 0
⇒ (2 sin2 2x - 1) (2 sin2 2x - 15) = 0
Therefore,
either, 2 sin2 2x - 1 = 0 ……….(1) or, 2 sin2 2x - 15 = 0 …………(2)
Now, from (1) we get,
1 - 2 sin2 2x = 0
⇒ cos 4x = 0
⇒ 4x = (2n + 1)π2, where, n ∈ Z
⇒ x = (2n + 1)π8, where, n ∈ Z
Again, from (2) we get, 2 sin2 2x = 15
⇒ sin2 2x = 152 which is impossible, since the numerical value of sin 2x cannot be greater than 1.
Therefore, the required general solution is: x = (2n + 1)π8, where, n ∈ Z
11 and 12 Grade Math
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