Name: Trigonometric Equation to Convert Cosine or Sine
Uploaded: 2019-07-19
Duration: 4 min 11 s
Description: Watch the video to learn how to solve trigonometric equations involving both sine and cosine simultaneously to find the general solution of the given equation.

Question

Solve the following equation:

\sin a x + \cos b x = 0

Accepted Answer

Given $\sin (a x) + \cos (b x) = 0$

$\Rightarrow \sin (a x) + \sin (\frac{π}{2} - b x) = 0$

$\Rightarrow 2 \sin (\frac{a x + \frac{π}{2} - b x}{2}) \cos (\frac{a x - \frac{π}{2} + b x}{2}) = 0$

$\Rightarrow 2 \sin ((a - b) \frac{x}{2} + \frac{π}{4}) \cos ((a + b) \frac{x}{2} - \frac{π}{4}) = 0$

Case - I:

$\sin ((a - b) \frac{x}{2} + \frac{π}{4}) = 0$

$\Rightarrow \sin ((a - b) \frac{x}{2} + \frac{π}{4}) = \sin (n π)$ , $n \in ℤ$

$\Rightarrow (a - b) \frac{x}{2} + \frac{π}{4} = n π$ , $n \in ℤ$

$\Rightarrow (a - b) \frac{x}{2} = n π - \frac{π}{4}$ , $n \in ℤ$

$\Rightarrow x = \frac{2}{(a - b)} (n π - \frac{π}{4}), n \in ℤ$

Case - II:

$\cos ((a + b) \frac{x}{2} - \frac{π}{4}) = 0$

$\Rightarrow \cos ((a + b) \frac{x}{2} - \frac{π}{4}) = \cos (2 n + 1) \frac{π}{2}$ , $n \in ℤ$

$\Rightarrow (a + b) \frac{x}{2} - \frac{π}{4} = (2 n + 1) \frac{π}{2}$ , $n \in ℤ$

$\Rightarrow (a + b) \frac{x}{2} = (2 n + 1) \frac{π}{2} + \frac{π}{4}$ , $n \in ℤ$

$\Rightarrow (a + b) x = (2 n + 1) π + \frac{π}{2}$ , $n \in ℤ$

$\Rightarrow x = \frac{1}{a + b} ((2 n + 1) π + \frac{π}{2})$ , $n \in ℤ$

K. C. Sinha Solutions for Chapter: Trigonometric Equations, Exercise 1: EXERCISE 5.1

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha Solutions for Chapter: Trigonometric Equations, Exercise 1: EXERCISE 5.1

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