Tricky Trigonometry

write the equation as

=> 2 * cos x = 1 - sin x

square both sides and use the identity cos ^ x + sin ^ x=1

you will get a quadratic in sin x which is of the form

=> 5 * sin ^ x - 2 * sin x - 3 = 0

solving this, we obtain the value of sin ^ x = 1 ; -3/5

find cos x and put the value in the given equation.

At fst find the value of sinx 2cosx=1-sinx after squaring and factorising(5sinx+3)(sinx-1)=0. So sinx=1,-3/5 and so ans is 2 and 6

2.cos x + sin x = 1

OR, 6(2.cos x + sin x) = 6

or, 12.cos x + 6.sin x = 6

or ,6.sin x = 6 - 12.cos x.....(1)

2.cos x + sin x = 1

or, sin x = 1 -2.cos x

Squaring both sides,we get

4.cos^2 x + 1 – 4.cos x = sin^2 x

Or, 4.cos^2 x + 1 – 4.cos x = 1 – cos^2 x

Or ,5.cos^2 x -4.cos x = 0

or,cos x(5.cos x - 4) = 0

Therefore.either cos x =0 0r 4/5

So, 7.cos x + 6.sin x = 6 - 5.cos x ...............(from 1)

if cos x =0,then 6 - 5.cos x = 6

if cos x = 4/5 , 6 - 5.cos x = 2