1.

Solve 4 cos2 θ = 3 (0° < θ < 360°)

Answer»

4 cos2θ = 3 

⇒ cos2 θ= \(\frac{3}{4}\) 

⇒ cos θ= \(\pm \frac{\sqrt{3}}{2}\)

cos θ = \( \frac{\sqrt{3}}{2}\) ⇒ θ = 30°, 330° (∵ cos θ is +ve and so θ lies in 1st and 4th quad.) 

cos θ = \(- \frac{\sqrt{3}}{2}\)  ⇒ θ = 150°, 210° (∵ cos θ is –ve and so θ lies in 2nd and 3rd quad.) 

⇒ θ = 30°, 150°, 210°, 330°. 

Note: if the value of θ is α when θ lies in the Ist quadrant then it is 180° – α
, 180° + α and 360° – α, If θ lies in 2nd, 3rd, and 4th quadrant respectively.



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