Solve for theta, cos theta + √3 sin theta = √2 , 0° < theta < 360°
Answers
Answer:
\bold{\theta=30^{\circ}}θ=30
∘
is the value of \bold{\sqrt{3} \sin \theta-\cos \theta=0}
3
sinθ−cosθ=0 if \bold{0 < \theta < 90^{\circ}.}0<θ<90
∘
.
Given:
\sqrt{3} \sin \theta-\cos \theta=0
3
sinθ−cosθ=0
0 < \theta < 90^{\circ}.0<θ<90
∘
.
To find:
Value of θ =?
Solution:
The question is \sqrt{3} \sin \theta-\cos \theta=0
3
sinθ−cosθ=0
Now to solve the question we transfer the cos θ on the other side of the equal to with which we get
\sqrt{3} \sin \theta-\cos \theta=0
3
sinθ−cosθ=0
\sqrt{3}=\frac{\cos \theta}{\sin \theta}
3
=
sinθ
cosθ
\sqrt{3}=\cot \theta
3
=cotθ
Therefore, transferring the cot on the other side of the equal to we get the inverse value of the cot i.e.
\cot ^{-1} \sqrt{3}=\thetacot
−1
3
=θ
Hence, the value of θ is proved to be 30, now the question says the θ is between zero degree and 90 degree thereby, proving that the value of \bold{\theta=30^{\circ}.}θ=30
∘
.