2cos^2x=sinx+1; 0° less than equal to x less than equal to 360°

2cos2xsinx1

#1

2cos^2x=sinx+1; 0° less than equal to x less than equal to 360°

Answer:

2cos^2x = sinx + 1
=>2(1-sin^2x) = sinx + 1 (Using the trigonometric identity, cos^2x + sin^2x = 1)
=>2 - 2sin^2x = sinx + 1
=>2sin^2x+sinx-1 = 0 ;
(This is a quadratic in sinx; factorising the equation or can solve for roots of the equation )
=>2sin^2x+2sinx-sinx-1 = 0
=>2sinx(sinx+1)-(sinx+1) = 0
=>(2sinx-1)(sinx+1)=0
=>sinx = 1/2 or sinx=-1;
sinx=1/2 => x=30 degrees or x=180-30 = 150 degrees ;
sinx=-1 => x= 180+90 = 270 degrees