Advertisements
Advertisements
प्रश्न
Solve: 2cos2θ + sin θ - 2 = 0.
Advertisements
उत्तर
2cos2θ + sin θ - 2 = 0
⇒ 2( 1 - sin2θ) + sin θ - 2 = 0
⇒ 2 - 2 sin2θ + sin θ - 2 = 0
⇒ - sin θ( 2 sin θ - 1) = 0
⇒ sin θ( 2 sin θ - 1) = 0
⇒ sin θ = 0 or 2 sin θ - 1 = 0
⇒ sin θ = 0 or sin θ = `1/2`
⇒ θ = 30°
संबंधित प्रश्न
What is the value of (cos2 67° – sin2 23°)?
Show that : `sin26^circ/sec64^circ + cos26^circ/(cosec64^circ) = 1`
Use tables to find the acute angle θ, if the value of tan θ is 0.2419
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
If \[\sec\theta = \frac{13}{12}\], find the values of other trigonometric ratios.
Given
\[\frac{4 \cos \theta - \sin \theta}{2 \cos \theta + \sin \theta}\] what is the value of \[\frac{{cosec}^2 \theta - \sec^2 \theta}{{cosec}^2 \theta + \sec^2 \theta}\]
The value of cos2 17° − sin2 73° is
Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°
If x and y are complementary angles, then ______.
If x tan 60° cos 60°= sin 60° cot 60°, then x = ______.
