If (2cos 2A-1) (tan 3A -1) = 0: find all the possible values of A.

Open in App

Solution

(2cos 2A -1) (tan 3A -1) = 0.
Hence either (2cos 2A - 1 = 0 or (tan 3A - 1) = 0
2 cos 2A - 1 = 0
$c o s 2 A = \frac{1}{2}$
$2 A = \frac{π}{3}, o r A = \frac{π}{3}, 2 n π \frac{+}{-} \frac{π}{3} .$
tan 3A - 1 = 0
tan 3A = 1
$3 A = \frac{π}{4}$
$A = \frac{π}{12}, n π \frac{+}{-} \frac{π}{12}$
Hence all the values of A are 2n $π \frac{+}{-} \frac{π}{3} a n d n π \frac{+}{-} \frac{π}{12}$

Suggest Corrections

Similar questions

(2 cos 2 A - 1) (tan 3 A - 1) = 0

A

30^{\circ}

A

∠ A

2 {cos}^{2} A - 3 cos A + 1 = 0

2 c o s^{2} A - 3 c o s A + 1 = 0

0^{\circ}

If 49 A and A49, where A > 0, have a common factor, find all possible values of A.

If ${lim}_{x \to a} \frac{x^{3} - a^{3}}{x - a} = {lim}_{x \to 1} \frac{x^{4} - 1}{x - 1}$ , find all possible values of a.

Join BYJU'S Learning Program