If log 0-5sin x-1-log 0-5cos x then number of values of x -2 - 2 - isA- 1B- 2C- 3D- 4

The correct option is D 4log_0.5 sinx = 1 log_0.5 cosx x ∈ [ 2π ,2π ] sinx gt; 0 and cosx gt; 0 sin 2x = 1 2x ∈ [ 4π ,4π ] ⇒4 solutions ...

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General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2

If log 0.5sin...

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If log $_{0. 5}$ sinx = 1 - log $_{0. 5}$ cosx then number of values of $x \in [- 2 π, 2 π]$ is

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{log}_{0.5} sin x = 1 - {log}_{0.5} cos x

x \in [- 2 π, 2 π]

{log}_{0.5} sin x = 1 - {log}_{0.5} cos x

x \in [- 2 π, 2 π]

{log}_{0.5} sin x = 1 - {log}_{0.5} cos x

x

- 2 π \leq x \leq 2 π

{log}_{0.5} sin x = 1 - {log}_{0.5} cos x

[- 2 π, 2 π]

{log}_{0.5} cos x = 1 - {log}_{0.5} sin x

x \in [- 2 π, 2 π]

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