2√2 sinΘ +2=0
Solve the equation for 0≤θ< 2π
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2√2 sinΘ +2=0
Solve the equation for 0≤θ< 2π
Obviously, Sin theta = -1/sqrt(2)
Now, we know that Sin 45 = 1/sqrt(2)
Now use the following facts: Sin is negative in third and fourth quadrants
Hence Sin (180 + A) = - Sin A and Sin (360 - A) = - Sin A
2root2 sin@=-2
sin@= -2/2root2
sin@= -1/root2
now look on to th trignometeric table and find out the value when sin@=-1/root2
plzzz tell this helped or not
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