To solve the equation 2sin²𝜃 = -sin𝜃, we first rewrite it as 2sin²𝜃 + sin𝜃 = 0.

This is a quadratic equation in sin𝜃. We can factor it as sin𝜃(2sin𝜃 + 1) = 0.

Setting each factor equal to zero gives the solutions sin𝜃 = 0 and sin𝜃 = -1/2.

The solutions to sin𝜃 = 0 in the interval 0≤𝜃

Question

To solve the equation 2sin²𝜃 = -sin𝜃, we first rewrite it as 2sin²𝜃 + sin𝜃 = 0.

This is a quadratic equation in sin𝜃. We can factor it as sin𝜃(2sin𝜃 + 1) = 0.

Setting each factor equal to zero gives the solutions sin𝜃 = 0 and sin𝜃 = -1/2.

The solutions to sin𝜃 = 0 in the interval 0≤𝜃<2⁢𝜋 are 𝜃 = 0 and 𝜃 = 𝜋.

The solutions to sin𝜃 = -1/2 in the interval 0≤𝜃<2⁢𝜋 are 𝜃 = 7𝜋/6 and 𝜃 = 11𝜋/6.

So, the solutions to the original equation are 𝜃 = 0, 𝜃 = 𝜋, 𝜃 = 7𝜋/6, and 𝜃 = 11𝜋/6.

Knowee AI · Accepted Answer

To solve the equation 2sin²𝜃 = -sin𝜃, we first rewrite it as 2sin²𝜃 + sin𝜃 = 0.

This is a quadratic equation in sin𝜃. We can factor it as sin𝜃(2sin𝜃 + 1) = 0.

Setting each factor equal to zero gives the solutions sin𝜃 = 0 and sin𝜃 = -1/2.

The solutions to sin𝜃 = 0 in the interval 0≤𝜃<2⁢𝜋 are 𝜃 = 0 and 𝜃 = 𝜋.

The solutions to sin𝜃 = -1/2 in the interval 0≤𝜃<2⁢𝜋 are 𝜃 = 7𝜋/6 and 𝜃 = 11𝜋/6.

So, the solutions to the original equation are 𝜃 = 0, 𝜃 = 𝜋, 𝜃 = 7𝜋/6, and 𝜃 = 11𝜋/6.

What are all values of 𝜃, for 0≤𝜃<2⁢𝜋, where 2⁢sin2⁡𝜃=-sin⁡𝜃 ?