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Solve 2sin^2http://www.ana.pro.br/bloggerimages/sonho2706.jpg -7sin http://www.ana.pro.br/bloggerimages/sonho2706.jpg= -3 for http://www.ana.pro.br/bloggerimages/sonho2706.jpg, where 0°< http://www.ana.pro.br/bloggerimages/sonho2706.jpg < 360°.
How would I do this???:confused:
2 sin^2(theta) - 7 sin(theat) = -3Quote:
Originally Posted by superman
is a quadratic in sin(theta). Solve this quadratic using to quadratic formula
to get sin(theta) = 1/2 or 3. Now sin(theta)=3 has no solution for theta in
the given range, so we are left with:
sin(theta) = 1/2, for theta in (0, 360] degrees, which you should be able to do
RonL
Hello, superman!
Quote:
Solve: .2·sin²θ -7·sin θ .= .-3 .for 0°< θ < 360°.
We have a quadratic equation: .2·sin²θ - 7·sin θ + 3 .= .0
. . which factors: .(sin θ - 3)(2·sin θ - 1) .= .0
And we have two equations to solve:
. . sin θ - 3 .= .0 . → . sin θ .= .3 . . . . no real solutions
. . 2·sin θ - 1 .= .0 . → . sin θ = ½ . → . θ .= .30°, 150°