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.

cos 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

• General solution of the equation sin x = ½
• General solution of the equation cos x = 1/√2
• General solution of the equation tan x = √3
• General Solution of the Equation sin θ = 0
• General Solution of the Equation cos θ = 0
• General Solution of the Equation tan θ = 0
• General Solution of the Equation sin θ = sin ∝
  
• General Solution of the Equation sin θ = 1
• General Solution of the Equation sin θ = -1
• General Solution of the Equation cos θ = cos ∝
• General Solution of the Equation cos θ = 1
• General Solution of the Equation cos θ = -1
• General Solution of the Equation tan θ = tan ∝
• General Solution of a cos θ + b sin θ = c
• Trigonometric Equation Formula
• Trigonometric Equation using Formula
• General solution of Trigonometric Equation
• Problems on Trigonometric Equation

11 and 12 Grade Math

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