First slide

Trigonometric equations

Question

General solution of $\sin^{2} x - 5 \sin xcos x - 6 \cos^{2} x = 0$ is

Moderate

A

$x = nπ - π / 4, n \in Z only$
B

$nπ + \tan^{- 1} 6, n \in Z only$
C

both (1) and (2)
D

none of these

Solution

Dividing the given equation by $\cos^{2} x, as \cos x = 0$ does not satisfy the equation, we have
$\begin{array}{l} \tan^{2} x - 5 \tan x - 6 = 0 \\ or (\tan x + 1) (\tan x - 6) = 0 \\ or \tan x = - 1 or \tan x = 6 \\ if \tan x = - 1 = \tan (- π / 4), then \\ x = nπ - π / 4, \forall n \in Z \end{array}$
and, if tanx = 6 = tan $α$ , (say)
$\begin{array}{l} \Rightarrow α = \tan^{- 1} 6, then \\ x = nπ + α = nπ + \tan^{- 1} 6, \forall n \in Z \\ Hence, x = nπ - (π / 4), nπ + \tan^{- 1} 6, n \in Z . \end{array}$

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