Question

find the value of theta, if 2cos^2theta +Sintheta-
2=0​

Answer 1

Answer:

Θ=0°,60°

Step-by-step explanation:

2$cos^{2}$Θ+sinΘ-2=0

cos2Θ+sinΘ-1=0(2$cos^{2}$Θ-1=cos2Θ)

1-2$sin^{2}$Θ+sinΘ-1=0(cos2Θ=1-2$sin^{2}$Θ)

sinΘ-2$sin^{2}$Θ=0

sinΘ(1-2sinΘ)=0

1. sinΘ=0       2.1-2sinΘ=0

  Θ=0°               2sinΘ=1

                          sinΘ=1/2

                          Θ=30°

Answer 2

Answer:

$\theta = 0° \\ \\ \theta = 30°$

Step-by-step explanation:

$2 \: {cos}^{2} \theta + sin \: \theta - 2 = 0$

it can be converted in quadratic equation in sine function

$2 \:(1 - {sin}^{2} \theta )+ sin \: \theta - 2 = 0 \\ \because \: {sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\ 2 \: - 2 {sin}^{2} \theta + sin \: \theta - 2 = 0 \\ \\ - 2 {sin}^{2} \theta + sin \: \theta = 0 \\ \\ sin \theta(1 - 2sin \theta) = 0 \\ \\ sin \theta = 0 \\ \\ \theta = 0° \\ or \\ \\ 1 - 2sin \theta = 0 \\ \\ sin\theta = \frac{1}{2} \\ \\ \theta = 30°$

Hope it helps you.