Solve 2 sin x+1<0 for 0 ≤ x<2 pi?
Update:
oh wait i think it is 2sin(x) + 1 < 0
it does not really say but i think thats it
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oh wait i think it is 2sin(x) + 1 < 0
it does not really say but i think thats it
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Verified answer
Please clarify...is this
2sin(x+1) < 0
or
2sin(x) + 1 < 0
OK, let's go with
2sin(x) + 1 < 0
2sin(x) < -1
sin x < -½
7π/6 < x < 11π/6
Understanding why is easy if you think about how sine looks on the unit circle. Sine is positive in Quadrants I & II, then becomes negative in Quadrants III & IV. Magnitude increases from as x goes from π to 3π/2, then decreases again as x goes to 2π.