How do you find the exact value of 2cos^2theta-3sintheta=0 in the interval 0<=theta<360?
1 Answer
May 10, 2018
Explanation:
"using the "color(blue)"trigonometric identity"
•color(white)(x)sin^2x+cos^2x=1
rArrcos^2x=1-sin^2x
rArr2(1-sin^2theta)-3sintheta=0
rArr2-2sin^2theta-3sintheta=0
rArr2sin^2theta+3sintheta-2=0
"we have a quadratic in sine which factors as"
(2sintheta-1)(sintheta+2)=0
"equate each factor to zero and solve for "theta
sintheta+2=0larrcolor(blue)"has no solution"
"since "-1<=sintheta<=1
2sintheta-1=0rArrsintheta=1/2
"since "sintheta>0" then "theta" in first/second quadrant"
theta=sin^-1(1/2)=30^@larrcolor(red)"first quadrant"
"or "theta=(180-30)^@=150^@larrcolor(red)"second quad"
rArrtheta=30^@" or "theta=150^@to(0<=theta<=360)
