Question

Solve the values of x in sin(x-32)=0

for 0° is less than or equal to x and is less than or equal to 180°​

Answer 1

$\huge\mathcal{\underline{\color{aqua} Good\: Afternoon}}$

$\huge{\orange{\boxed{\fcolorbox{lime}{cadetblue}{\pink{ANSWER}}}}}$

$\red\checkmark\:\sin(x - 32) = 0 \\ \\ = > \sin(x - 32) = \sin0 \\ \\ = > x - 32 = 0 \\ \\ = > x = 32$

⭐ Therefore, the value of x is 32 .

Answer 2

Question :-

Solve the values of x in sin(x-32)=0

for 0°

Answer :-

x = 32

Given :-

x in sin(x-32)=0

To Find :-

Value of x

Solution :-

For such kind of QuEsTiOn first we have to know the value of trigonometry angles at 0° , 30° , 45° , 60° and 90°

Now , as per the Question :-

As we all know the ratios now ,

sin ( x - 32 ) = 0°

{ Transposing sin } we get :-

( x - 32 ) = sin 0 ( sin = 0 by substituting the value of sin 0° from the trigonometry table )

( x - 32 ) = 0

Now ,

x = 32 ( by transposing { - 32 }

More to know

{ Formulas }

1. tan θ = sin θ / cos θ

2. cossecθ = 1/ sinθ

3. sec θ = 1/cosθ

4. Cotθ = 1/ tanθ

5. Sin²θ+ Cos²θ= 1

6. Sec²θ - tan²θ = 1

7. cosec ²θ - cot²θ = 1

8. sin(90°−θ) = cos θ

9. cos(90°−θ) = sin θ

10. tan(90°−θ) = cot θ

11. cot(90°−θ) = tan θ

12. sec(90°−θ) = cosec θ

13. cosec(90°−θ) = sec θ

14. Sin2θ = 2 sinθ cosθ

15. cos2θ = Cos²θ- Sin²θ

16. cot θ = cos θ / sin θ.

For Trigonometry table refer to the attachment.