Question

given that sec x - 5 tan x = 3 cos x , show that 3 sin2 x - 5 sin x - 2 = 0

Answer 1

secx-5tanx=3cosx or secx/cosx-5tanx/cosx-3=0 or 1/cos^2x-5sinx/cos^2x-3=0 or 1-5sinx-3cos^2x=0 or 1-5sinx-3(1-sin^2x)=0 or 1-5sinx-3+3sin^2x=0 or3sin^2x-5sinx-2=0 ...hence proved.

Answer 2

Answer:

It has been proves that:

$3\,sin^2x-5sinx-2=0$.

Step-by-step explanation:

Given that:

$secx-5tanx=3cosx$

Dividing both side of the given equation by cosx.

Hence we get:

$\dfrac{secx}{cosx}-\dfrac{5tanx}{cosx}=\dfrac{3cosx}{cosx}$

Formula used below has given here:

$secx=\dfrac{1}{cosx}\\tanx=\dfrac{sinx}{cosx}$

Therefore:

$\dfrac{1}{cosx\cdot cosx}-\dfrac{\dfrac{5sinx}{cosx}}{cosx}=3$

$\dfrac{1}{cos^2x}-\dfrac{5sinx}{cos^2x}=3$

$\dfrac{1-5sinx}{cos^2x}=3$

$1-5sinx=3cos^2x$

Using trigonometric formulas:

$sin^2x+cos^2x=1\\cos^2x=(1-sin^2x)$

Substituting the values:

$1-5sinx=3(1-sin^2x)\\1-5sinx=3-3sin^2x\\3sin^2x+1-3-5sinx=0\\3sin^2x-5sinx-2=0$

Proved.