If 7sin2x+3cos2x=4 show that tanx=1/√3
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7 sin² x + 3 cos ² x = 4
7 sin² x + 3 (1- sin²x) = 4 (since cos² x = 1 - sin² x)
7 sin² x+3 - 3 sin² x = 4
4 sin² x = 1
sin² x = 1/4
sin x = 1/2
sin x = sin 30 °
x = 30°
LHS
tan x = tan 30°
= 1/√3 = RHS
7 sin² x + 3 (1- sin²x) = 4 (since cos² x = 1 - sin² x)
7 sin² x+3 - 3 sin² x = 4
4 sin² x = 1
sin² x = 1/4
sin x = 1/2
sin x = sin 30 °
x = 30°
LHS
tan x = tan 30°
= 1/√3 = RHS
Answered by
3
7 sin^2 x + 3 cos ^ 2 x = 4
7 sin ^2 x + 3 ( 1 - sin^2 x) = 4 ::::(cos^2 x = 1 - sin^2 x)
7 sin ^ 2 x + 3 - 3 sin ^2 x = 4
4 sin ^ 2 x = 1
sin ^2 x = 1/4
sin x = 1/2
sin x = sin 30
x = 30
LHS tan = x = tan = 30
= 1/√3
hence proved LHS =RHS
7 sin ^2 x + 3 ( 1 - sin^2 x) = 4 ::::(cos^2 x = 1 - sin^2 x)
7 sin ^ 2 x + 3 - 3 sin ^2 x = 4
4 sin ^ 2 x = 1
sin ^2 x = 1/4
sin x = 1/2
sin x = sin 30
x = 30
LHS tan = x = tan = 30
= 1/√3
hence proved LHS =RHS
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