trigonometric equations
plz solve
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Hey dear ☺️
Here your answer
2sin²@ = 3cos@
2(1-cos²@) = 3cos@
2-2cos²@= 3cos@
2cos²@+3cos@-2=0
2cos²+4cos@-cos@-2=0
2cos@( cos@+ 2) - 1(cos@+2)=0
(2cos-1)(cos@+2)=0
2cos@-1=0 or cos@+2=0
Cos@= 1/2 or cos@= - 2
But cos@=-2 is not possible
& cos@= 1/2 is possible
Cos@= 1/2
@= cos¯1/2
@= π/3
Hope it helpful
Here your answer
2sin²@ = 3cos@
2(1-cos²@) = 3cos@
2-2cos²@= 3cos@
2cos²@+3cos@-2=0
2cos²+4cos@-cos@-2=0
2cos@( cos@+ 2) - 1(cos@+2)=0
(2cos-1)(cos@+2)=0
2cos@-1=0 or cos@+2=0
Cos@= 1/2 or cos@= - 2
But cos@=-2 is not possible
& cos@= 1/2 is possible
Cos@= 1/2
@= cos¯1/2
@= π/3
Hope it helpful
samrocks1234:
thank you
:-)
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