In the interval -(π)/2,(π)/2, the equation log_(sin⁡θ)⁡cos⁡2θ=2 has

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) a unique solution. Since, (-π)/2≤θ≤(π)/2⇒sin⁡θ∈-1,1 Here sin⁡θ∈0,1 for log_(sin⁡θ)⁡cos⁡2θ=2 to be defined. ⇒cos⁡2θ=sin^2⁡θ ⇒1-2sin^2⁡θ=sin^2⁡θ ⇒sin^2⁡θ=1/3⇒sin⁡θ=1/(√3) (as sin⁡θ∈0,1 ) ⇒The equation has a unique solution.

In the interval $[- \frac{π}{2}, \frac{π}{2}]$ , the equation $\log_{\sin θ} (\cos 2 θ) = 2$ has

Solution

More Trigonometric Equations Questions

In the interval -π2,π2, the equation logsin⁡θ⁡cos⁡2θ=2 has

Solution

More Trigonometric Equations Questions

In the interval $[- \frac{π}{2}, \frac{π}{2}]$ , the equation $\log_{\sin θ} (\cos 2 θ) = 2$ has