Help needed in understanding the process to solve a trig problem such as [ 2sin^2x-3x= 0 ]
Remember the relation between sine and cosine? sine^2+cosine^2=1, so sin^2X=1-cos^2X. Substitute for sin^2X:
We now have a quadratic equation. If it helps, let y= cosX, then:
2-2y^2-3y=0 or 2y^2+3y-2=0, which factorises:
(2y-1)(y+2)=0, and y=1/2 or -2 which means cosX=1/2 or -2. But we know that cosX can't be bigger than 1 or less than -1, so we can safely reject cosX=-2, and solve cosX=1/2. This is one of those familiar angles, X=60 degrees or (pi)/3 radians. Now you need to remember which quadrants cosine is positive in. ASTC (All Silver Tea Cups: All trigs, sin, tan, cos) for Q1 to 4. So cosine is positive in Q1 and Q4. The solution therefore is (pi)/3 and 2(pi)-(pi)/3=5(pi)/3 or 60 and 360-60=300 degrees. Read More: ...