Given : $$2cos\theta=2-sin\theta$$
=> $$2cos\theta=2-\sqrt{1-cos^2\theta}$$
=> $$\sqrt{1-cos^2\theta}=2-2cos\theta$$
Squaring both sides,
=> $$(\sqrt{1-cos^2\theta})^2=(2-2cos\theta)^2$$
=> $$1-cos^2\theta=4+4cos^2\theta-8cos\theta$$
Let $$cos\theta=x$$
=> $$5x^2-8x+3=0$$
=> $$5x^2-5x-3x+3=0$$
=> $$5x(x-1)-3(x-1)=0$$
=> $$(5x-3)(x-1)=0$$
=> $$x=cos\theta=1,\frac{3}{5}$$
=> Ans - (A)
Create a FREE account and get: