Question

Find all the angles between 0 and 360 degrees which satisfy the equations:

(i) sin x + 3cos x =0
(ii) sin (2y + 60 degrees )= -0.5

Answer 1

Let's solve each equation individually:

(i) \( \sin x + 3 \cos x = 0 \)

Rearranging, we get \( \sin x = -3 \cos x \). Divide both sides by \( \cos x \) (assuming \( \cos x \\eq 0 \)), and we get \( \tan x = -3 \).

So, \( x = \arctan(-3) \) is one solution. To find other solutions, we can add \( 180^\circ \) to \( x \) because \( \tan(x + 180^\circ) = \tan x \).

(ii) \( \sin(2y + 60^\circ) = -0.5 \)

This implies \( 2y + 60^\circ = \arcsin(-0.5) \). Solve for \( y \) by isolating \( y \).

After finding \( y \) from equation (ii), check if \( 0 \leq y < 360^\circ \).

Combine the solutions from both equations to find the common values of \( x \) and \( y \) that satisfy both equations.

Answer 2

Explanation:

(i)

`\boxed{a\cdot sin(x)+b\cdot cos(x)=R\cdot sin(x+\alpha)}`

`where\ R=√(a^2+b^2)\ and\ \alpha=tan^(-1)((b)/(a));\ 0^o\leq \alpha\leq 90^o`

sin x + 3cos x = 0

`R=√(1^2+3^2)`

`=√(10)`

`\alpha=tan^(-1)((3)/(1) )`

`=71.57^o`

`√(10) \cdot sin(x+71.57^o)=0`

`sin(x+71.57^o)=0`

`x+71.57^o=k\cdot 180^o`

`x=k\cdot 180^o-71.57^o`

`x=\{108.43^o, 288.43^o\}`

(ii)

sin(2y+60°)=-0.5

2y + 60° = (210° + k·360°) or (330° + k·360°)

2y = (150° + k·360°) or (270° + k·360°)

y = (75° + k·180°) or (135° + k·180°)

y = {75°, 135°, 255°, 315°}