Question

Pls help solve for x...thank you

(3/cos^2x) - 7 - (4/cot x) = 0

Answer 1

\[\frac{1}{\cos^2 x}=1+\tan^2 x \\ and \\ \frac{1}{\cot x}=\tan x\]

Answer 2

\[\frac{3}{\cos^2 x} - 7 - \frac{4}{\frac{\cos x}{\sin x}} = 0\]
\[\frac{3}{\cos^2 x} - 7 - \frac{4\sin x}{\cos x} = 0\]
\[\frac{3}{\cos^2 x} - \frac{4\sin x}{\cos x} - 7 = 0\]
\[3 \sec^2 x - 4 \tan x - 7 = 0\]
\[3 (\tan^2 x + 1) - 4\tan x - 7 = 0\]
\[3\tan^2 x + 3 - 4 \tan x -7 = 0\]
\[3\tan^2 x - 4\tan x - 10 = 0\]
let a = tan x
\[3a^2 - 4a - 10 = 0\]
\[(3a +5)(a - 3) = 0\]
\[ a = -5/3 \; ; \; a = 3\]
\[\tan x = -5/3 \;; \; \tan x = 3\]
\[x = \tan^{-1} (- 5/3) \; ;\; \tan^{-1} (3)\]

Answer 3

do you get that @kwokyy ?

Answer 4

Thanks guys...